Question

$\frac{8}{\sqrt[3]{(2x^{2}-7x-5)^{11}}}$

Answer 1

8(2x^{2}-7x-5)^{-11/3}

Answer 2

The given expression is $\frac{8}{\sqrt[3]{(2x^{2}-7x-5)^{11}}}$.

Step 1: Convert the cube root into a fractional exponent.

We know that $\sqrt[n]{A} = A^{1/n}$. So, $\sqrt[3]{(2x^{2}-7x-5)^{11}} = ((2x^{2}-7x-5)^{11})^{1/3}$.

Step 2: Apply the power of a power rule, $(a^m)^n = a^{mn}$.

$((2x^{2}-7x-5)^{11})^{1/3} = (2x^{2}-7x-5)^{11 \times \frac{1}{3}} = (2x^{2}-7x-5)^{\frac{11}{3}}$.

Step 3: Rewrite the expression with the simplified denominator.

The expression becomes $\frac{8}{(2x^{2}-7x-5)^{\frac{11}{3}}}$.

Step 4: Use the rule $\frac{1}{a^n} = a^{-n}$ to move the term from the denominator to the numerator.

$\frac{8}{(2x^{2}-7x-5)^{\frac{11}{3}}} = 8 \times (2x^{2}-7x-5)^{-\frac{11}{3}}$.

The simplified expression is $8(2x^{2}-7x-5)^{-11/3}$.