Question

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Answer 1

Answer:

$3x^3$

Step-by-step explanation:

$\left(\frac{81x^{12}}{3x^3}\right)^{\frac{1}{3}}$

$\mathrm{Apply\:exponent\:rule}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}$

$=\frac{\left(81x^{12}\right)^{\frac{1}{3}}}{\left(3x^3\right)^{\frac{1}{3}}}$

$=\frac{3\cdot \:3^{\frac{1}{3}}x^4}{\left(3x^3\right)^{\frac{1}{3}}}$

$=\frac{3\cdot \:3^{\frac{1}{3}}x^4}{3^{\frac{1}{3}}x}$

$=\frac{3^{\frac{1}{3}+1}x^4}{3^{\frac{1}{3}}x}$

$\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}$

$\frac{3^{\frac{1}{3}+1}}{3^{\frac{1}{3}}}=3^{1+\frac{1}{3}-\frac{1}{3}}$

$=\frac{3^{\frac{1}{3}-\frac{1}{3}+1}x^4}{x}$

$\mathrm{Subtract\:the\:numbers:}\:1+\frac{1}{3}-\frac{1}{3}=1$

$=\frac{3x^4}{x}$

$=3x^3$