How do you solve 3x^(4/3) + 5 = 53 and find any extraneous solutions?

Aug 10, 2016

$x = 8$

Explanation:

$3 {x}^{\frac{4}{3}} + 5 = 53$
or
$3 {x}^{\frac{4}{3}} = 53 - 5$
or
$3 {x}^{\frac{4}{3}} = 48$
or
${x}^{\frac{4}{3}} = \frac{48}{3}$
or
${x}^{\frac{4}{3}} = 16$
or
$x = {16}^{\frac{3}{4}}$
or
$x = \sqrt[4]{{16}^{3}}$
or
$x = {\left(\sqrt[4]{2 \times 2 \times 2 \times 2}\right)}^{3}$
or
$x = {2}^{3}$
or
$x = 8$