Evaluate x^3+40=25

$x^{3} + 40 = 25$

Move all terms not containing $x$ to the right side of the equation.

Subtract $40$ from both sides of the equation.

$x^{3} = 25 - 40$

Subtract $40$ from $25$ .

$x^{3} = - 15$

Take the cube root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{- 15}$

Simplify $\sqrt[3]{- 15}$ .

Rewrite $- 15$ as ${(- 1)}^{3} \cdot 15$ .

Rewrite $- 15$ as $- 1 (15)$ .

$x = \sqrt[3]{- 1 (15)}$

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$x = \sqrt[3]{{(- 1)}^{3} \cdot 15}$

Pull terms out from under the radical.

$x = - 1 \sqrt[3]{15}$

Rewrite $- 1 \sqrt[3]{15}$ as $- \sqrt[3]{15}$ .

$x = - \sqrt[3]{15}$

The result can be shown in multiple forms.

Exact Form:

$x = - \sqrt[3]{15}$

Decimal Form:

$x = - 2.46621207 \dots$