{"id":147,"date":"2016-06-01T18:10:16","date_gmt":"2016-06-01T18:10:16","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=147"},"modified":"2019-08-06T19:00:37","modified_gmt":"2019-08-06T19:00:37","slug":"special-cases-sum-and-difference-of-cubes","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/special-cases-sum-and-difference-of-cubes\/","title":{"raw":"Special Cases - Cubes","rendered":"Special Cases &#8211; Cubes"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Factor special products<\/li>\r\n<\/ul>\r\n<\/div>\r\nSome interesting patterns arise when you are working with cubed quantities within polynomials. Specifically, there are two more special cases to consider: [latex]a^{3}+b^{3}[\/latex] and [latex]a^{3}-b^{3}[\/latex].\r\n\r\nLet us take a look at how to factor sums and differences of cubes.\r\n<h2>Sum of Cubes<\/h2>\r\nThe term \u201ccubed\u201d is used to describe a number raised to the third power. In geometry, a cube is a six-sided shape with equal width, length, and height; since all these measures are equal, the volume of a cube with width\u00a0[latex]x[\/latex] can be represented by [latex]x^{3}[\/latex].\u00a0(Notice the exponent!)\r\n\r\nCubed numbers get large very quickly: [latex]1^{3}=1[\/latex], [latex]2^{3}=8[\/latex], [latex]3^{3}=27[\/latex], [latex]4^{3}=64[\/latex], and [latex]5^{3}=125[\/latex]\r\n\r\nBefore looking at factoring a sum of two cubes, let us look at the possible factors.\r\n\r\nIt turns out that [latex]a^{3}+b^{3}[\/latex] can actually be factored as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex]. Check these factors by multiplying.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nDoes [latex](a+b)(a^{2}\u2013ab+b^{2})=a^{3}+b^{3}[\/latex]?\r\n[reveal-answer q=\"386615\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"386615\"]\r\n\r\nApply the distributive property.\r\n<p style=\"text-align: center;\">[latex]\\left(a\\right)\\left(a^{2}\u2013ab+b^{2}\\right)+\\left(b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex]<\/p>\r\nMultiply by <i>a<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\left(a^{3}\u2013a^{2}b+ab^{2}\\right)+\\left(b\\right)\\left(a^{2}-ab+b^{2}\\right)[\/latex]<\/p>\r\nMultiply by <i>b<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\left(a^{3}\u2013a^{2}b+ab^{2}\\right)+\\left(a^{2}b\u2013ab^{2}+b^{3}\\right)[\/latex]<\/p>\r\nRearrange terms in order to combine the like terms.\r\n<p style=\"text-align: center;\">[latex]a^{3}-a^{2}b+a^{2}b+ab^{2}-ab^{2}+b^{3}[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center;\">[latex]a^{3}+b^{3}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nDid you see that? Four of the terms cancelled out, leaving us with the (seemingly) simple binomial [latex]a^{3}+b^{3}[\/latex]. So, the factors are correct.\r\n\r\nYou can use this pattern to factor binomials in the form [latex]a^{3}+b^{3}[\/latex], otherwise known as \u201cthe sum of cubes.\u201d\r\n<div class=\"textbox shaded\">\r\n<h3>The Sum of Cubes<\/h3>\r\nA binomial in the form [latex]a^{3}+b^{3}[\/latex] can be factored as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex].\r\n<h3>Examples<\/h3>\r\nThe factored form of [latex]x^{3}+64[\/latex]\u00a0is [latex]\\left(x+4\\right)\\left(x^{2}\u20134x+16\\right)[\/latex].\r\n\r\nThe factored form of [latex]8x^{3}+y^{3}[\/latex] is [latex]\\left(2x+y\\right)\\left(4x^{2}\u20132xy+y^{2}\\right)[\/latex].\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]x^{3}+8y^{3}[\/latex].\r\n[reveal-answer q=\"354149\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"354149\"]\r\n\r\nIdentify that this binomial fits the sum of cubes pattern [latex]a^{3}+b^{3}[\/latex].\r\n\r\n[latex]a=x[\/latex], and [latex]b=2y[\/latex] (since [latex]2y\\cdot2y\\cdot2y=8y^{3}[\/latex]).\r\n<p style=\"text-align: center;\">[latex]x^{3}+8y^{3}[\/latex]<\/p>\r\nFactor the binomial as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex], substituting [latex]a=x[\/latex] and [latex]b=2y[\/latex] into the expression.\r\n<p style=\"text-align: center;\">[latex]\\left(x+2y\\right)\\left(x^{2}-x\\left(2y\\right)+\\left(2y\\right)^{2}\\right)[\/latex]<\/p>\r\nSquare\u00a0[latex](2y)^{2}=4y^{2}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\left(x+2y\\right)\\left(x^{2}-x\\left(2y\\right)+4y^{2}\\right)[\/latex]<\/p>\r\nMultiply [latex]\u2212x\\left(2y\\right)=\u22122xy[\/latex] (writing the coefficient first).\r\n<p style=\"text-align: left;\">The factored form is [latex]\\left(x+2y\\right)\\left(x^{2}-2xy+4y^{2}\\right)[\/latex].<\/p>\r\n<p style=\"text-align: left;\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nNow try another one.\r\n\r\nYou should always look for a common factor before you follow any of the patterns for factoring.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]16m^{3}+54n^{3}[\/latex].\r\n[reveal-answer q=\"254227\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"254227\"]\r\n\r\nFactor out the common factor\u00a0[latex]2[\/latex].\r\n<p style=\"text-align: center;\">[latex]16m^{3}+54n^{3}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]2\\left(8m^{3}+27n^{3}\\right)[\/latex]<\/p>\r\n[latex]8m^{3}[\/latex] and [latex]27n^{3}[\/latex]\u00a0are cubes, so you can factor [latex]8m^{3}+27n^{3}[\/latex]\u00a0as the sum of two cubes: [latex]a=2m[\/latex] and\u00a0[latex]b=3n[\/latex].\r\n\r\nFactor the binomial [latex]8m^{3}+27n^{3}[\/latex] substituting [latex]a=2m[\/latex]\u00a0and [latex]b=3n[\/latex] into the expression [latex]\\left(a+b\\right)\\left(a^{2}-ab+b^{2}\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]2\\left(2m+3n\\right)\\left[\\left(2m\\right)^{2}-\\left(2m\\right)\\left(3n\\right)+\\left(3n\\right)^{2}\\right][\/latex]<\/p>\r\nSquare: [latex](2m)^{2}=4m^{2}[\/latex] and [latex](3n)^{2}=9n^{2}[\/latex].\r\n<p style=\"text-align: center;\">[latex]2\\left(2m+3n\\right)\\left[4m^{2}-\\left(2m\\right)\\left(3n\\right)+9n^{2}\\right][\/latex]<\/p>\r\nMultiply [latex]-\\left(2m\\right)\\left(3n\\right)=-6mn[\/latex].\r\n<p style=\"text-align: left;\">The factored form is [latex]2\\left(2m+3n\\right)\\left(4m^{2}-6mn+9n^{2}\\right)[\/latex].<\/p>\r\n<p style=\"text-align: left;\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<h2>Difference of Cubes<\/h2>\r\nHaving seen how binomials in the form [latex]a^{3}+b^{3}[\/latex]\u00a0can be factored, it should not come as a surprise that binomials in the form [latex]a^{3}-b^{3}[\/latex]\u00a0can be factored in a similar way.\r\n<div class=\"textbox shaded\">\r\n<h3>The Difference of Cubes<\/h3>\r\nA binomial in the form [latex]a^{3}\u2013b^{3}[\/latex] can be factored as [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex].\r\n<h3>Examples<\/h3>\r\nThe factored form of [latex]x^{3}\u201364[\/latex]\u00a0is [latex]\\left(x\u20134\\right)\\left(x^{2}+4x+16\\right)[\/latex].\r\n\r\nThe factored form of [latex]27x^{3}\u20138y^{3}[\/latex] is [latex]\\left(3x\u20132y\\left)\\right(9x^{2}+6xy+4y^{2}\\right)[\/latex].\r\n\r\n<\/div>\r\nNotice that the basic construction of the factorization is the same as it is for the sum of cubes; the difference is in the [latex]+[\/latex] and [latex]\u2013[\/latex] signs. Take a moment to compare the factored form of [latex]a^{3}+b^{3}[\/latex]\u00a0with the factored form of [latex]a^{3}-b^{3}[\/latex].\r\n\r\nFactored form of\u00a0[latex]a^{3}+b^{3}[\/latex]:\u00a0[latex]\\left(a+b\\right)\\left(a^{2}-ab+b^{2}\\right)[\/latex]\r\n\r\nFactored form of [latex]a^{3}-b^{3}[\/latex]: [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex]\r\n\r\nThis can be tricky to remember because of the different signs. The factored form of [latex]a^{3}+b^{3}[\/latex]\u00a0contains a negative, and the factored form of [latex]a^{3}-b^{3}[\/latex]<i> <\/i>contains a positive! Some people remember the different forms like this:\r\n\r\n\u201cRemember one sequence of variables: [latex]a^{3}b^{3}=\\left(a\\,b\\right)\\left(a^{2}ab\\,b^{2}\\right)[\/latex].\u00a0There are\u00a0[latex]4[\/latex] missing signs. Whatever the first sign is, it is also the second sign. The third sign is the opposite, and the fourth sign is always [latex]+[\/latex].\u201d\r\n\r\nTry this for yourself. If the first sign is [latex]+[\/latex], as in [latex]a^{3}+b^{3}[\/latex], according to this strategy, how do you fill in the rest: [latex]\\left(a\\,b\\right)\\left(a^{2}ab\\,b^{2}\\right)[\/latex]? Does this method help you remember the factored form of\u00a0[latex]a^{3}+b^{3}[\/latex]\u00a0and\u00a0[latex]a^{3}\u2013b^{3}[\/latex]?\r\n\r\nLet us go ahead and look at a couple of examples. Remember to factor out all common factors first.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]8x^{3}\u20131,000[\/latex].\r\n[reveal-answer q=\"809204\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"809204\"]\r\n\r\nFactor out\u00a0[latex]8[\/latex].\r\n<p style=\"text-align: center;\">[latex]8(x^{3}\u2013125)[\/latex]<\/p>\r\nIdentify that the binomial fits the pattern\u00a0[latex]a^{3}-b^{3}:a=x[\/latex], and [latex]b=5[\/latex] (since [latex]5^{3}=125[\/latex]).\r\n\r\nFactor [latex]x^{3}\u2013125[\/latex]\u00a0as [latex]\\left(a\u2013b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex], substituting [latex]a=x[\/latex] and [latex]b=5[\/latex] into the expression.\r\n<p style=\"text-align: center;\">[latex]8\\left(x-5\\right)\\left[x^{2}+\\left(x\\right)\\left(5\\right)+5^{2}\\right][\/latex]<\/p>\r\nSquare the first and last terms, and rewrite\u00a0[latex]\\left(x\\right)\\left(5\\right)[\/latex] as [latex]5x[\/latex].\r\n<p style=\"text-align: center;\">[latex]8\\left(x\u20135\\right)\\left(x^{2}+5x+25\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nHere is one more example. Note that [latex]r^{9}=\\left(r^{3}\\right)^{3}[\/latex]\u00a0and that [latex]8s^{6}=\\left(2s^{2}\\right)^{3}[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]r^{9}-8s^{6}[\/latex].\r\n[reveal-answer q=\"609560\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"609560\"]\r\n\r\nIdentify this binomial as the difference of two cubes. As shown above, it is.\r\n\r\nRewrite [latex]r^{9}[\/latex] as [latex]\\left(r^{3}\\right)^{3}[\/latex] and rewrite [latex]8s^{6}[\/latex] as [latex]\\left(2s^{2}\\right)^{3}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\left(r^{3}\\right)^{3}-\\left(2s^{2}\\right)^{3}[\/latex]<\/p>\r\nNow the binomial is written in terms of cubed quantities. Thinking of [latex]a^{3}-b^{3}[\/latex], [latex]a=r^{3}[\/latex] and [latex]b=2s^{2}[\/latex].\r\n\r\nFactor the binomial as [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex], substituting [latex]a=r^{3}[\/latex] and [latex]b=2s^{2}[\/latex] into the expression.\r\n<p style=\"text-align: center;\">[latex]\\left(r^{3}-2s^{2}\\right)\\left[\\left(r^{3}\\right)^{2}+\\left(r^{3}\\right)\\left(2s^{2}\\right)+\\left(2s^{2}\\right)^{2}\\right][\/latex]<\/p>\r\nMultiply and square the terms.\r\n<p style=\"text-align: center;\">[latex]\\left(r^{3}-2s^{2}\\right)\\left(r^{6}+2r^{3}s^{2}+4s^{4}\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following two video examples, we show more binomials that can be factored as a sum or difference of cubes.\r\n\r\nhttps:\/\/youtu.be\/tFSEpOB262M\r\n\r\nhttps:\/\/youtu.be\/J_0ctMrl5_0\r\n\r\nYou encounter some interesting patterns when factoring. Two special cases\u2014the sum of cubes and the difference of cubes\u2014can help you factor some binomials that have a degree of three (or higher, in some cases). The special cases are:\r\n<ul>\r\n \t<li>A\u00a0binomial in the form [latex]a^{3}+b^{3}[\/latex]\u00a0can be factored as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex]<\/li>\r\n \t<li>A binomial in the form [latex]a^{3}-b^{3}[\/latex]\u00a0can be factored as [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex]<\/li>\r\n<\/ul>\r\nAlways remember to factor out any common factors first.","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Factor special products<\/li>\n<\/ul>\n<\/div>\n<p>Some interesting patterns arise when you are working with cubed quantities within polynomials. Specifically, there are two more special cases to consider: [latex]a^{3}+b^{3}[\/latex] and [latex]a^{3}-b^{3}[\/latex].<\/p>\n<p>Let us take a look at how to factor sums and differences of cubes.<\/p>\n<h2>Sum of Cubes<\/h2>\n<p>The term \u201ccubed\u201d is used to describe a number raised to the third power. In geometry, a cube is a six-sided shape with equal width, length, and height; since all these measures are equal, the volume of a cube with width\u00a0[latex]x[\/latex] can be represented by [latex]x^{3}[\/latex].\u00a0(Notice the exponent!)<\/p>\n<p>Cubed numbers get large very quickly: [latex]1^{3}=1[\/latex], [latex]2^{3}=8[\/latex], [latex]3^{3}=27[\/latex], [latex]4^{3}=64[\/latex], and [latex]5^{3}=125[\/latex]<\/p>\n<p>Before looking at factoring a sum of two cubes, let us look at the possible factors.<\/p>\n<p>It turns out that [latex]a^{3}+b^{3}[\/latex] can actually be factored as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex]. Check these factors by multiplying.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Does [latex](a+b)(a^{2}\u2013ab+b^{2})=a^{3}+b^{3}[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q386615\">Show Solution<\/span><\/p>\n<div id=\"q386615\" class=\"hidden-answer\" style=\"display: none\">\n<p>Apply the distributive property.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(a\\right)\\left(a^{2}\u2013ab+b^{2}\\right)+\\left(b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex]<\/p>\n<p>Multiply by <i>a<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(a^{3}\u2013a^{2}b+ab^{2}\\right)+\\left(b\\right)\\left(a^{2}-ab+b^{2}\\right)[\/latex]<\/p>\n<p>Multiply by <i>b<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(a^{3}\u2013a^{2}b+ab^{2}\\right)+\\left(a^{2}b\u2013ab^{2}+b^{3}\\right)[\/latex]<\/p>\n<p>Rearrange terms in order to combine the like terms.<\/p>\n<p style=\"text-align: center;\">[latex]a^{3}-a^{2}b+a^{2}b+ab^{2}-ab^{2}+b^{3}[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center;\">[latex]a^{3}+b^{3}[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>Did you see that? Four of the terms cancelled out, leaving us with the (seemingly) simple binomial [latex]a^{3}+b^{3}[\/latex]. So, the factors are correct.<\/p>\n<p>You can use this pattern to factor binomials in the form [latex]a^{3}+b^{3}[\/latex], otherwise known as \u201cthe sum of cubes.\u201d<\/p>\n<div class=\"textbox shaded\">\n<h3>The Sum of Cubes<\/h3>\n<p>A binomial in the form [latex]a^{3}+b^{3}[\/latex] can be factored as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex].<\/p>\n<h3>Examples<\/h3>\n<p>The factored form of [latex]x^{3}+64[\/latex]\u00a0is [latex]\\left(x+4\\right)\\left(x^{2}\u20134x+16\\right)[\/latex].<\/p>\n<p>The factored form of [latex]8x^{3}+y^{3}[\/latex] is [latex]\\left(2x+y\\right)\\left(4x^{2}\u20132xy+y^{2}\\right)[\/latex].<\/p>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]x^{3}+8y^{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q354149\">Show Solution<\/span><\/p>\n<div id=\"q354149\" class=\"hidden-answer\" style=\"display: none\">\n<p>Identify that this binomial fits the sum of cubes pattern [latex]a^{3}+b^{3}[\/latex].<\/p>\n<p>[latex]a=x[\/latex], and [latex]b=2y[\/latex] (since [latex]2y\\cdot2y\\cdot2y=8y^{3}[\/latex]).<\/p>\n<p style=\"text-align: center;\">[latex]x^{3}+8y^{3}[\/latex]<\/p>\n<p>Factor the binomial as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex], substituting [latex]a=x[\/latex] and [latex]b=2y[\/latex] into the expression.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+2y\\right)\\left(x^{2}-x\\left(2y\\right)+\\left(2y\\right)^{2}\\right)[\/latex]<\/p>\n<p>Square\u00a0[latex](2y)^{2}=4y^{2}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+2y\\right)\\left(x^{2}-x\\left(2y\\right)+4y^{2}\\right)[\/latex]<\/p>\n<p>Multiply [latex]\u2212x\\left(2y\\right)=\u22122xy[\/latex] (writing the coefficient first).<\/p>\n<p style=\"text-align: left;\">The factored form is [latex]\\left(x+2y\\right)\\left(x^{2}-2xy+4y^{2}\\right)[\/latex].<\/p>\n<p style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<p>Now try another one.<\/p>\n<p>You should always look for a common factor before you follow any of the patterns for factoring.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]16m^{3}+54n^{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q254227\">Show Solution<\/span><\/p>\n<div id=\"q254227\" class=\"hidden-answer\" style=\"display: none\">\n<p>Factor out the common factor\u00a0[latex]2[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]16m^{3}+54n^{3}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]2\\left(8m^{3}+27n^{3}\\right)[\/latex]<\/p>\n<p>[latex]8m^{3}[\/latex] and [latex]27n^{3}[\/latex]\u00a0are cubes, so you can factor [latex]8m^{3}+27n^{3}[\/latex]\u00a0as the sum of two cubes: [latex]a=2m[\/latex] and\u00a0[latex]b=3n[\/latex].<\/p>\n<p>Factor the binomial [latex]8m^{3}+27n^{3}[\/latex] substituting [latex]a=2m[\/latex]\u00a0and [latex]b=3n[\/latex] into the expression [latex]\\left(a+b\\right)\\left(a^{2}-ab+b^{2}\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]2\\left(2m+3n\\right)\\left[\\left(2m\\right)^{2}-\\left(2m\\right)\\left(3n\\right)+\\left(3n\\right)^{2}\\right][\/latex]<\/p>\n<p>Square: [latex](2m)^{2}=4m^{2}[\/latex] and [latex](3n)^{2}=9n^{2}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]2\\left(2m+3n\\right)\\left[4m^{2}-\\left(2m\\right)\\left(3n\\right)+9n^{2}\\right][\/latex]<\/p>\n<p>Multiply [latex]-\\left(2m\\right)\\left(3n\\right)=-6mn[\/latex].<\/p>\n<p style=\"text-align: left;\">The factored form is [latex]2\\left(2m+3n\\right)\\left(4m^{2}-6mn+9n^{2}\\right)[\/latex].<\/p>\n<p style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<h2>Difference of Cubes<\/h2>\n<p>Having seen how binomials in the form [latex]a^{3}+b^{3}[\/latex]\u00a0can be factored, it should not come as a surprise that binomials in the form [latex]a^{3}-b^{3}[\/latex]\u00a0can be factored in a similar way.<\/p>\n<div class=\"textbox shaded\">\n<h3>The Difference of Cubes<\/h3>\n<p>A binomial in the form [latex]a^{3}\u2013b^{3}[\/latex] can be factored as [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex].<\/p>\n<h3>Examples<\/h3>\n<p>The factored form of [latex]x^{3}\u201364[\/latex]\u00a0is [latex]\\left(x\u20134\\right)\\left(x^{2}+4x+16\\right)[\/latex].<\/p>\n<p>The factored form of [latex]27x^{3}\u20138y^{3}[\/latex] is [latex]\\left(3x\u20132y\\left)\\right(9x^{2}+6xy+4y^{2}\\right)[\/latex].<\/p>\n<\/div>\n<p>Notice that the basic construction of the factorization is the same as it is for the sum of cubes; the difference is in the [latex]+[\/latex] and [latex]\u2013[\/latex] signs. Take a moment to compare the factored form of [latex]a^{3}+b^{3}[\/latex]\u00a0with the factored form of [latex]a^{3}-b^{3}[\/latex].<\/p>\n<p>Factored form of\u00a0[latex]a^{3}+b^{3}[\/latex]:\u00a0[latex]\\left(a+b\\right)\\left(a^{2}-ab+b^{2}\\right)[\/latex]<\/p>\n<p>Factored form of [latex]a^{3}-b^{3}[\/latex]: [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex]<\/p>\n<p>This can be tricky to remember because of the different signs. The factored form of [latex]a^{3}+b^{3}[\/latex]\u00a0contains a negative, and the factored form of [latex]a^{3}-b^{3}[\/latex]<i> <\/i>contains a positive! Some people remember the different forms like this:<\/p>\n<p>\u201cRemember one sequence of variables: [latex]a^{3}b^{3}=\\left(a\\,b\\right)\\left(a^{2}ab\\,b^{2}\\right)[\/latex].\u00a0There are\u00a0[latex]4[\/latex] missing signs. Whatever the first sign is, it is also the second sign. The third sign is the opposite, and the fourth sign is always [latex]+[\/latex].\u201d<\/p>\n<p>Try this for yourself. If the first sign is [latex]+[\/latex], as in [latex]a^{3}+b^{3}[\/latex], according to this strategy, how do you fill in the rest: [latex]\\left(a\\,b\\right)\\left(a^{2}ab\\,b^{2}\\right)[\/latex]? Does this method help you remember the factored form of\u00a0[latex]a^{3}+b^{3}[\/latex]\u00a0and\u00a0[latex]a^{3}\u2013b^{3}[\/latex]?<\/p>\n<p>Let us go ahead and look at a couple of examples. Remember to factor out all common factors first.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]8x^{3}\u20131,000[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q809204\">Show Solution<\/span><\/p>\n<div id=\"q809204\" class=\"hidden-answer\" style=\"display: none\">\n<p>Factor out\u00a0[latex]8[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]8(x^{3}\u2013125)[\/latex]<\/p>\n<p>Identify that the binomial fits the pattern\u00a0[latex]a^{3}-b^{3}:a=x[\/latex], and [latex]b=5[\/latex] (since [latex]5^{3}=125[\/latex]).<\/p>\n<p>Factor [latex]x^{3}\u2013125[\/latex]\u00a0as [latex]\\left(a\u2013b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex], substituting [latex]a=x[\/latex] and [latex]b=5[\/latex] into the expression.<\/p>\n<p style=\"text-align: center;\">[latex]8\\left(x-5\\right)\\left[x^{2}+\\left(x\\right)\\left(5\\right)+5^{2}\\right][\/latex]<\/p>\n<p>Square the first and last terms, and rewrite\u00a0[latex]\\left(x\\right)\\left(5\\right)[\/latex] as [latex]5x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]8\\left(x\u20135\\right)\\left(x^{2}+5x+25\\right)[\/latex]<\/p>\n<p style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<p>Here is one more example. Note that [latex]r^{9}=\\left(r^{3}\\right)^{3}[\/latex]\u00a0and that [latex]8s^{6}=\\left(2s^{2}\\right)^{3}[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]r^{9}-8s^{6}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q609560\">Show Solution<\/span><\/p>\n<div id=\"q609560\" class=\"hidden-answer\" style=\"display: none\">\n<p>Identify this binomial as the difference of two cubes. As shown above, it is.<\/p>\n<p>Rewrite [latex]r^{9}[\/latex] as [latex]\\left(r^{3}\\right)^{3}[\/latex] and rewrite [latex]8s^{6}[\/latex] as [latex]\\left(2s^{2}\\right)^{3}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\left(r^{3}\\right)^{3}-\\left(2s^{2}\\right)^{3}[\/latex]<\/p>\n<p>Now the binomial is written in terms of cubed quantities. Thinking of [latex]a^{3}-b^{3}[\/latex], [latex]a=r^{3}[\/latex] and [latex]b=2s^{2}[\/latex].<\/p>\n<p>Factor the binomial as [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex], substituting [latex]a=r^{3}[\/latex] and [latex]b=2s^{2}[\/latex] into the expression.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(r^{3}-2s^{2}\\right)\\left[\\left(r^{3}\\right)^{2}+\\left(r^{3}\\right)\\left(2s^{2}\\right)+\\left(2s^{2}\\right)^{2}\\right][\/latex]<\/p>\n<p>Multiply and square the terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(r^{3}-2s^{2}\\right)\\left(r^{6}+2r^{3}s^{2}+4s^{4}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following two video examples, we show more binomials that can be factored as a sum or difference of cubes.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 1:  Factor a Sum or Difference of Cubes\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/tFSEpOB262M?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 3:  Factor a Sum or Difference of Cubes\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/J_0ctMrl5_0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You encounter some interesting patterns when factoring. Two special cases\u2014the sum of cubes and the difference of cubes\u2014can help you factor some binomials that have a degree of three (or higher, in some cases). The special cases are:<\/p>\n<ul>\n<li>A\u00a0binomial in the form [latex]a^{3}+b^{3}[\/latex]\u00a0can be factored as [latex]\\left(a+b\\right)\\left(a^{2}\u2013ab+b^{2}\\right)[\/latex]<\/li>\n<li>A binomial in the form [latex]a^{3}-b^{3}[\/latex]\u00a0can be factored as [latex]\\left(a-b\\right)\\left(a^{2}+ab+b^{2}\\right)[\/latex]<\/li>\n<\/ul>\n<p>Always remember to factor out any common factors first.<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-147\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Authored by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li><strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/tFSEpOB262M\">https:\/\/youtu.be\/tFSEpOB262M<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Unit 12: Factoring, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 3: Factor a Sum or Difference of Cubes. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/J_0ctMrl5_0\">https:\/\/youtu.be\/J_0ctMrl5_0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/tFSEpOB262M\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 12: Factoring, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of 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