How do you write the simplified form of $- 64^{\frac{1}{3}}$ $-64^(1/ 3)$ ?

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How do you write the simplified form of $- 64^{\frac{1}{3}}$ $-64^(1/ 3)$ ?

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Answer 1 · 2018-04-24T04:24:43

the simplified answer would be -4

Explanation:

Let's factor out 64:
$64 = 2^{6}$ $64=2^6$
$- {(2^{6})}^{\frac{1}{3}}$ $-(2^6)^(1/3)$
$= - 2^{6 . (\frac{1}{3})}$ $=-2^(6.(1/3))$
$= - 2^{2}$ $=-2^2$
$= - 4$ $=-4$

Answer 2 · 2018-04-24T06:35:58

$- 4$ $-4$

Explanation:

Recall one of the laws of indices:

$\sqrt{x} = x^{\frac{1}{2}} and \sqrt[3]{x} = x^{\frac{1}{3}}$ $sqrtx = x^(1/2)" "and " "root3(x) = x^(1/3)$

$- 64^{\frac{1}{3}} = \sqrt[3]{- 64}$ $-64^(1/3) = root3(-64)$

$64$ $64$ is a perfect cube: $64 = 4^{3}$ $64=4^3$

$\sqrt[3]{- 64} = - 4$ $root3(-64) =-4$

You could also work with the prime factors:

$\sqrt[3]{- 64} = \sqrt[3]{- (2^{6})}$ $root3(-64) = root3(-(2^6))$

$= - 2^{2}$ $=-2^2$

$= - 4$ $=-4$

Note that perfect cubes can be negative, but perfect squares cannot.

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How do you write the simplified form of $- 64^{\frac{1}{3}}$ $-64^(1/ 3)$ ?

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How do you write the simplified form of $- 64^{\frac{1}{3}}$ $-64^(1/ 3)$ ?