Question

The value of 3' x10, when x = -2, can be

Answer 1

Answer:

$a = 8$ and $b = 2$

Step-by-step explanation:

Given

$\sqrt[3]{x^{10}}$

$x = -2$

Required

Express as: $a\sqrt[3]{b}$

Substitute -2 for x in $\sqrt[3]{x^{10}}$

$\sqrt[3]{x^{10}} = \sqrt[3]{(-2)^{10}}$

$\sqrt[3]{x^{10}} = \sqrt[3]{1024}$

Express 1024 as 2^10

$\sqrt[3]{x^{10}} = \sqrt[3]{2^{10}}$

Apply law of indices:

$\sqrt[3]{x^{10}} = \sqrt[3]{2^{9+1}}$

Apply law of indices: Split

$\sqrt[3]{x^{10}} = \sqrt[3]{2^{9}*2^1}}$

$\sqrt[3]{x^{10}} = \sqrt[3]{2^{9}} *\sqrt[3]{2^1}}$

$\sqrt[3]{x^{10}} = \sqrt[3]{2^{9}} *\sqrt[3]{2}}$

$\sqrt[3]{x^{10}} = 2^{9*\frac{1}{3}}} *\sqrt[3]{2}}$

$\sqrt[3]{x^{10}} = 2^3 *\sqrt[3]{2}}$

$\sqrt[3]{x^{10}} = 8\sqrt[3]{2}}$

By comparison:

$a = 8$ and $b = 2$