1. The two radicals have the same radicand and index. This means you can combine them by adding “coefficients,” like in a polynomial.

   [latex] \text{3}\sqrt{11}+ 7\sqrt{11}=10\sqrt{11}[/latex]

2. Rearrange terms so that like terms are next to each other. Then add.

   [latex] 5\sqrt{2}+2\sqrt{2}+\sqrt{3}+4\sqrt{3}=7\sqrt{2}+5\sqrt{3}[/latex]

   We cannot simplify any further because the remaining terms have different radicands.

3. The radicands and indices are the same, so these two radical terms can be combined.

   [latex] 5\sqrt{13}-3\sqrt{13}=2\sqrt{13}[/latex]

4. Rearrange terms so that like terms are next to each other. Then add.

   [latex] 3\sqrt{x}-7\sqrt{x}+12\sqrt[3]{x}=-4\sqrt{x}+12\sqrt[3]{x}[/latex]

   This is the answer since the two remaining terms have different indices.

1. First remove all perfect square factors from the radicals, then combine like terms.

   [latex]\begin{align}\sqrt{28}-3\sqrt{63}&=\sqrt{4\cdot7}-3\cdot\sqrt{9\cdot7}\\ &=\sqrt{2^2}\cdot\sqrt{7}-3\cdot\sqrt{3^2}\cdot\sqrt{7}\\ &=2\sqrt{7}-3\cdot 3\cdot\sqrt{7}\\ &=2\sqrt{7}-9\sqrt{7}\\ &=-7\sqrt{7}\end{align}[/latex]

2. First remove all perfect cubes from the radicals, then combine like terms.

   [latex] \begin{align}2\sqrt[3]{40}+\sqrt[3]{135}&=2\sqrt[3]{8\cdot 5}+\sqrt[3]{27\cdot 5}\\ &=2\sqrt[3]{{{2}^{3}}}\cdot \sqrt[3]{5}+\sqrt[3]{{{3}^{3}}}\cdot \sqrt[3]{5}\\ &=2\cdot 2\cdot \sqrt[3]{5}+3\cdot \sqrt[3]{5}\\ &=4\sqrt[3]{5}+3\sqrt[3]{5}\\ &=7\sqrt[3]{5}\end{align}[/latex]

3. First remove all perfect cubes from the radicals, then combine like terms.

   [latex]\begin{align}x\sqrt[3]{x{{y}^{4}}}+y\sqrt[3]{{{x}^{4}}y}&=x\sqrt[3]{x\cdot {{y}^{3}}\cdot y}+y\sqrt[3]{{{x}^{3}}\cdot x\cdot y}\\&=x\sqrt[3]{{{y}^{3}}}\cdot \sqrt[3]{xy}+y\sqrt[3]{{{x}^{3}}}\cdot \sqrt[3]{xy}\\&=xy\cdot \sqrt[3]{xy}+xy\cdot \sqrt[3]{xy}\\ &=2xy\sqrt[3]{xy}\end{align}[/latex]

4. Simplify each radical by identifying and pulling out powers of [latex]4[/latex].

   [latex]\begin{align}5\sqrt[4]{{{a}^{5}}b}-a\sqrt[4]{16ab}&=5\sqrt[4]{{{a}^{4}}\cdot a\cdot b}-a\sqrt[4]{{{2}^{4}}\cdot a\cdot b}\\&=5\cdot a\sqrt[4]{a\cdot b}-a\cdot 2\sqrt[4]{a\cdot b}\\&=5a\sqrt[4]{ab}-2a\sqrt[4]{ab}\\&=3a\sqrt[4]{ab}\end{align}[/latex]

Use the Distributive Property.

[latex] \sqrt{x}(3\sqrt{x}-5)=\sqrt{x}\cdot 3\sqrt{x}-\sqrt{x}\cdot 5[/latex]

Now multiply the radicals and simplify if possible:

[latex] \begin{align} =&\quad 3\sqrt{x^2}-5\sqrt{x} \\ =&\quad 3x-5\sqrt{x}\end{align}[/latex]

Use the Distributive Property to multiply each term within parentheses by [latex] 7\sqrt{x}[/latex].

[latex] 7\sqrt{x}\left( 2\sqrt{xy}+\sqrt{y} \right)=7\sqrt{x}\left( 2\sqrt{xy} \right)+7\sqrt{x}\left( \sqrt{y} \right)[/latex]

Apply the rules of multiplying radicals.

[latex] =7\cdot 2\sqrt{{{x}^{2}}y}+7\sqrt{xy}[/latex]

Pull out the perfect square factor.

[latex] =14x\sqrt{y}+7\sqrt{xy}[/latex]

This is the answer because the remaining radicals are not like terms.

Use the Distributive Property.

[latex] \sqrt[3]{a}\left( 2\sqrt[3]{{{a}^{2}}}-4\sqrt[3]{{{a}^{5}}}+8\sqrt[3]{{{a}^{8}}} \right)=\sqrt[3]{a}\left( 2\sqrt[3]{{{a}^{2}}} \right)-\sqrt[3]{a}\left( 4\sqrt[3]{{{a}^{5}}} \right)+\sqrt[3]{a}\left( 8\sqrt[3]{{{a}^{8}}} \right)[/latex]

Apply the rules of multiplying radicals.

[latex] \begin{align}&=2\sqrt[3]{a\cdot {{a}^{2}}}-4\sqrt[3]{a\cdot {{a}^{5}}}+8\sqrt[3]{a\cdot {{a}^{8}}}\\&=2\sqrt[3]{{{a}^{3}}}-4\sqrt[3]{{{a}^{6}}}+8\sqrt[3]{{{a}^{9}}}\end{align}[/latex]

Identify cubes in each of the radicals.

[latex] \begin{align}&=2\sqrt[3]{{{a}^{3}}}-4\sqrt[3]{{{\left( {{a}^{2}} \right)}^{3}}}+8\sqrt[3]{{{\left( {{a}^{3}} \right)}^{3}}}\\&=2a-4{{a}^{2}}+8{{a}^{3}}\end{align}[/latex]

Use the Distributive Property to multiply.

[latex]\begin{align}&F & &O & &I & &L \\ &2\sqrt{6}\cdot 3\sqrt{6} & &2\sqrt{6}\cdot \left( -2 \right) & &5\cdot 3\sqrt{6} & &5\cdot \left( -2 \right) \\ &6\sqrt{6\cdot 6} & &-4\sqrt{6} & &15\sqrt{6} & &-10 \\ &36 & &-4\sqrt{6} & &15\sqrt{6} & &-10 \end{align}[/latex]

Now, combine any like terms.

[latex] =26 + 11\sqrt{6}[/latex]

1. FOIL, and then simplify the radicals.[latex]\begin{align}\left(4\sqrt{2}-3\sqrt{7}\right)\left(3\sqrt{2}+\sqrt{7}\right)&=4\cdot3\sqrt{2}\sqrt{2}+4\sqrt{2}\sqrt{7}-3\cdot 3\sqrt{7}\sqrt{2}-3\sqrt{7}\sqrt{7}\\ &=12(2)+4\sqrt{14}-9\sqrt{14}-3(7)&&\color{blue}{\sqrt{a}\sqrt{a}=a}\\ &=24+4\sqrt{14}-9\sqrt{14}-21\\ &=3-5\sqrt{14}&&\color{blue}{\textsf{combine like terms}}\end{align}[/latex]
2. First rewrite the power as binomial multiplication. Then use FOIL and simplify. Alternatively, you could use the formula [latex]\left(a+b\right)^2=a^2+2ab+b^2.[/latex]
   [latex]\begin{align}\left(5-2\sqrt{x}\right)^2&=\left(5-2\sqrt{x}\right)\left(5-2\sqrt{x}\right)\\ &=25-5\cdot 2\sqrt{x}-5\cdot 2\sqrt{x}+4\sqrt{x}\sqrt{x}\\ &=25-10\sqrt{x}-10\sqrt{x}+4x&&\color{blue}{\sqrt{x}\sqrt{x}=x}\\ &=25-20\sqrt{x}+4x&&\color{blue}{\textsf{combine like terms}}\end{align}[/latex]
3. FOIL and then simplify.
   [latex]\begin{align}\left(4+\sqrt{11}\right)\left(4-\sqrt{11}\right)&=16-4\sqrt{11}+4\sqrt{11}-\sqrt{11}\sqrt{11}\\ &=16-4\sqrt{11}+4\sqrt{11}-11&&\color{blue}{\sqrt{a}\sqrt{a}=a}\\ &=5&&\color{blue}{\textsf{combine like terms}}\end{align}[/latex]

1. Since we can write [latex]4=2^2[/latex], this is a situation where the radicands share a common base.
   [latex]\begin{align}\sqrt[4]{2}\cdot\sqrt[5]{4}&=\sqrt[4]{2}\cdot\sqrt[5]{2^2}\\ &=2^{1/4}\cdot 2^{2/5}&&\color{blue}{\textsf{convert to exponent form}}\\ &=2^{\frac{1}{4}+\frac{2}{5}}&&\color{blue}{\textsf{product rule for exponents}}\\ &=2^{\frac{5}{20}+\frac{8}{20}}&&\color{blue}{\textsf{common denominators}}\\ &=2^{13/20}\\ &=\sqrt[20]{2^{13}}\end{align}[/latex]
2. The radicands are not equal so we convert to exponent form, get a common denominator, and then factor out the denominator.
   [latex]\begin{align}\sqrt[7]{ab^4}\cdot\sqrt[6]{a^5b}&=\left(ab^4\right)^{1/7}(a^5b)^{1/6}\\ &=\left(ab^4\right)^{6/42}\left(a^5b\right)^{7/42}&&\color{blue}{\textsf{common denominators on exponents}}\\ &=\left(\left(ab^4\right)^6\right)^{1/42}\left(\left(a^5b\right)^7\right)^{1/42}&&\color{blue}{\textsf{power rule of exponents}}\\ &=\left(\left(ab^4\right)^6\left(a^5b\right)^7\right)^{1/42}&&\color{blue}{\textsf{product to a power}}\\ &=\left(a^6b^{24}a^{35}b^7\right)^{1/42}&&\color{blue}{\textsf{product to a power rule, and power rule}}\\ &=\left(a^{41}b^{31}\right)^{1/42}\\ &=\sqrt[42]{a^{41}b^{31}}\end{align}[/latex]
3. This problem has a common radicand so we convert to exponents and then subtract the exponents.
   [latex]\begin{align}\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}&=\frac{\sqrt[3]{x^2}}{\sqrt[6]{x}}\\ &=\frac{x^{2/3}}{x^{1/6}}\\ &=x^{\frac{2}{3}-\frac{1}{6}}&&\color{blue}{\textsf{quotient rule of exponents}}\\ &=x^{\frac{4}{6}-\frac{1}{6}}&&\color{blue}{\textsf{common denominators}}\\ &=x^{\frac{3}{6}}\\ &=x^{\frac{1}{2}}\\ &=\sqrt{x}\end{align}[/latex]So, [latex]\left(\dfrac{f}{g}\right)(x)=\sqrt{x}.[/latex]