# How do you solve and check for extraneous solutions in root3(x-3) =1?

Aug 7, 2015

$x = 4$ with no extraneous solutions

#### Explanation:

Given $\sqrt[3]{x - 3} = 1$

Cube both sides:
$\textcolor{w h i t e}{\text{XXXX}}$$x - 3 = {1}^{3} = 1$

$\textcolor{w h i t e}{\text{XXXX}}$$x = 4$
Verify by substituting $4$ for $x$ in the original expression
$\textcolor{w h i t e}{\text{XXXX}}$$\sqrt[3]{4 - 3} = \sqrt[3]{1} = 1$