How do you simplify $(\sqrt[3]{- 162 x^{6}})$ $(root3(-162x^6))$ ?

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How do you simplify $(\sqrt[3]{- 162 x^{6}})$ $(root3(-162x^6))$ ?

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Answer 1 · 2015-10-03T17:56:26

$\sqrt[3]{- 162 x^{6}} = x^{2} (- 3 \sqrt[3]{6})$ $root(3)(-162x^6) = x^2(-3root(3)(6))$

Explanation:

We can take a minus out of the root, because $- 1 = {(- 1)}^{3}$ $-1 = (-1)^3$

$- \sqrt[3]{162 x^{6}}$ $-root(3)(162x^6)$

We know that $x^{6} = x^{2} \cdot x^{2} \cdot x^{2}$ $x^6 = x^2 * x^2 * x^2$ or $x^{6} = {(x^{2})}^{3}$ $x^6 = (x^2)^3$ , so we can take a $x^{2}$ $x^2$ out of the root

$- x^{2} \sqrt[3]{162}$ $-x^2root(3)(162)$

Lastly, all we have to do is factor the $162$ $162$ , so

$162 ∣ 2$ $162|2$
$081 ∣ 3$ $color(white)(0)81|3$
$027 ∣ 3$ $color(white)(0)27|3$
$009 ∣ 3$ $color(white)(00)9|3$
$003 ∣ 3$ $color(white)(00)3|3$
$001 ∣ 2 \cdot 3^{4}$ $color(white)(00)1|2*3^4$

So we can take a $3$ $3$ out of the root, leaving a $2 \cdot 3$ $2*3$ , that is, $6$ $6$ inside it. So

$\sqrt[3]{- 162 x^{6}} = x^{2} (- 3 \sqrt[3]{6})$ $root(3)(-162x^6) = x^2(-3root(3)(6))$

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How do you simplify $(\sqrt[3]{- 162 x^{6}})$ $(root3(-162x^6))$ ?

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