# How do you solve  5^3=(x+2)^3?

Sep 10, 2015

$x = 3$

#### Explanation:

${5}^{3} = {\left(x + 2\right)}^{3}$ => rewrite as:

${\left(x + 2\right)}^{3} - {5}^{3} = 0$ => factor by the difference of cubes formula:

$\left[\left(x + 2\right) - 5\right] \left[{\left(x + 2\right)}^{2} + 5 \left(x + 2\right) + 25\right] = 0$ => simplify:

$\left(x - 3\right) \left({x}^{2} + 9 x + 39\right) = 0$ => equate each bracket to zero:

$x - 3 = 0 \implies x = 3$

${x}^{2} + 9 x + 39 = 0$

Discriminant $= {b}^{2} - 4 a c = 81 - 156 = - 75 < 0$

Since the discriminant is negative the quadratic has no real solutions therefore the only valid solution is:

$x = 3$