How do you solve 8x^3+27=0?

Dec 6, 2016

-1.5

Explanation:

$8 {x}^{3} = - 27$

${x}^{3} = - \frac{27}{8}$

x = cube root of -27/8

or $x = {\left(- \frac{27}{8}\right)}^{\frac{1}{3}}$