# How do you solve root3(3x - 8) = 1  and find any extraneous solutions?

Jun 21, 2018

$x = 3$

There are no extraneous solutions.

#### Explanation:

$\sqrt[3]{3 x - 8} = 1$

${\left(\sqrt[3]{3 x - 8}\right)}^{3} = {1}^{3}$

$3 x - 8 = 1$

$3 x = 9$

$x = 3$

test for "extraneous solutions":

$\sqrt[3]{3 x - 8} = 1$

$\sqrt[3]{3 \left(3\right) - 8} = 1$

$\sqrt[3]{9 - 8} = 1$

$\sqrt[3]{1} = 1$

$1 = 1$ Checks.