# How do you solve (7x - 2)^(1/3) + (7x + 5)^(1/3) = 3?

Jul 30, 2016

$x = \frac{3}{7}$

#### Explanation:

Note that if $7 x = 3$ then:

${\left(7 x - 2\right)}^{\frac{1}{3}} + {\left(7 x + 5\right)}^{\frac{1}{3}} = {1}^{\frac{1}{3}} + {8}^{\frac{1}{3}} = 1 + 2 = 3$

So one solution is $x = \frac{3}{7}$

Since ${t}^{\frac{1}{3}}$ is strictly monotonically increasing, this will be the only Real solution.

graph{(y-(7x-2)^(1/3)-(7x+5)^(1/3))(y-3)=0 [-10.16, 9.84, -2.88, 7.12]}