Question: "Adding and subtracting radicals 324a4b336ab23"

High School
Algebra
Algebra I
Exponents and radicals

boitshupoO

Answered question

2021-09-08

Adding and subtracting radicals
$\frac{\sqrt[3]{324 a^{4} b^{3}}}{\sqrt[3]{6 a b^{2}}}$

Answer & Explanation

opsadnojD

Skilled2021-09-09Added 95 answers

Given:
$\frac{\sqrt[3]{324 a^{4} b^{3}}}{\sqrt[3]{6 a b^{2}}}$
Expanding the expression.
Using $\sqrt[n]{p} = p^{\frac{1}{n}}$
$\frac{\sqrt[3]{324 a^{4} b^{3}}}{\sqrt[3]{6 a b^{2}}} = \frac{{(324 a^{4} b^{3})}^{\frac{1}{3}}}{{(6 a b^{2})}^{\frac{1}{3}}}$
Using ${(p q)}^{m} = p^{m} q^{m}$
$\frac{{(324 a^{4} b^{3})}^{\frac{1}{3}}}{{(6 a b^{2})}^{\frac{1}{3}}} = \frac{{(324)}^{\frac{1}{3}} a^{\frac{4}{3}} b^{\frac{3}{3}}}{6^{\frac{1}{3}} a^{\frac{1}{3}} b^{\frac{2}{3}}}$
Simplification:
Using ${(\frac{p}{q})}^{m} = \frac{p^{m}}{q^{m}}$ and $\frac{p^{m}}{q^{n}} = p^{m - n}$ ,
$\frac{{(324)}^{\frac{1}{3}} a^{\frac{4}{3}} b^{\frac{3}{3}}}{6^{\frac{1}{3}} a^{\frac{1}{3}} b^{\frac{2}{3}}} = {(\frac{324}{6})}^{\frac{1}{3}} a^{\frac{4}{3} - \frac{1}{3}} b^{1 - \frac{2}{3}}$
$\frac{{(324)}^{\frac{1}{3}} a^{\frac{4}{3}} b^{\frac{3}{3}}}{6^{\frac{1}{3}} a^{\frac{1}{3}} b^{\frac{2}{3}}} = {(54)}^{\frac{1}{3}} a^{\frac{3}{3}} b^{\frac{1}{3}}$
$\frac{{(324)}^{\frac{1}{3}} a^{\frac{4}{3}} b^{\frac{3}{3}}}{6^{\frac{1}{3}} a^{\frac{1}{3}} b^{\frac{2}{3}}} = ($

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