Question

Simplify the expression

Answer 1

Nice job inputting the expression.

The cube root is a 1/3 power. The 4 in the denominator is a -4 power in the numerator.   When we have powers of powers we multiply them all together.  When we have a product to a power we have to raise each factor to the power.

We get to choose whether we want a fraction at the end or negative exponents.  Because of the constant 16 in the denominator I chose fraction.

$\dfrac{1}{( \sqrt[3]{8p^6})^4} = ( \sqrt[3]{8p^6})^{-4} = ( (8p^6)^{\frac 1 3})^{-4} =(8^{\frac 1 3})^{-4} p^{(6(-4)/3)} = 2^{-4} p^{-8} = \dfrac{1}{16p^8}$

Answer: 1/(16p⁸)

Answer 2

Step-by-step explanation:

$\frac{1}{(\sqrt[3]{(8p^{6})^{4} } } \\ \\ = \frac{1}{(\sqrt[3]{ ({2}^{3} p^{6})^{4} } } \\ \\ = \frac{1}{(\sqrt[3]{ ({2}^{3 \times 4} p^{6 \times 4}) } }\\ \\ = \frac{1}{(\sqrt[3]{ ({2}^{12} \times p^{24}) } } \\ \\ = \frac{1}{({ {2}^{12 \times \frac{1}{3} } \times p^{24 \times \frac{1}{3} }) } } \\ \\ = \frac{1}{({ {2}^{4} \times p^{8 }) } } \\ \\ \purple{ \boxed{ \bold{ \therefore \: \frac{1}{(\sqrt[3]{(8p^{6})^{4} } } = \frac{1}{{16 p^{8 } } }}}}$