# How do you solve root3(x^2)= 9?

Mar 30, 2017

Undo each of the things done to $x$ one at a time.

#### Explanation:

First, to undo the cube root, we can cube both sides to get

${x}^{2} = 729$

Next, to undo the square, we can take the square root of both sides to get

$x = \pm 27$

Mar 30, 2017

$x = \pm 27$

#### Explanation:

Given:

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

Note that $9 = {3}^{2}$, so this can also be written:

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

Note that both $t \rightarrow {t}^{3}$ and its inverse $t \rightarrow \sqrt[3]{t}$ are one to one as functions of real numbers. So if we cube both sides of the equation then we neither lose any solution nor introduce any extraneous solutions:

Cubing both sides of the equation, we get:

${x}^{2} = {\left({3}^{2}\right)}^{3} = {3}^{2 \cdot 3} = {3}^{3 \cdot 2} = {\left({3}^{3}\right)}^{2} = {27}^{2}$

Subtract ${27}^{2}$ from both ends to get:

${x}^{2} - {27}^{2} = 0$

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Using this with $a = x$ and $b = 27$ we find:

$0 = {x}^{2} - {27}^{2} = \left(x - 27\right) \left(x + 27\right)$

So:

$x = \pm 27$