How do you solve $\sqrt[3]{2 x^{3} - 24} = - x$ $root3(2x^3 - 24) = -x$ and find any extraneous solutions?

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Answer 1 · 2016-06-27T14:04:15

Cube both sides.

${(\sqrt[3]{2 x^{3} - 24})}^{3} = {(- x)}^{3}$ $(root(3)(2x^3 - 24))^3 = (-x)^3$

$2 x^{3} - 24 = - x^{3}$ $2x^3 - 24 = -x^3$

$3 x^{3} = 24$ $3x^3 = 24$

$x^{3} = 8$ $x^3 = 8$

$x = 2$ $x = 2$

Checking in the original equation:

$\sqrt[3]{2 {(2)}^{3} - 24} =^{?} - 2$ $root(3)(2(2)^3 - 24 )=^? -2$

$\sqrt[3]{- 8} = - 2$ $root(3)(-8) = -2$

Therefore, out solution set is ${x = 2}$ ${x = 2}$ .

Hopefully this helps!

How do you solve root3(2x^3 - 24) = -x3√2x3−24=−xroot3(2x^3 - 24) = -x and find any extraneous solutions?