Solve [x^3/3+x^2-3x]_-4^-3

$$[ \frac { x ^ { 3 } } { 3 } + x ^ { 2 } - 3 x ] _ { - 4 } ^ { - 3 }$$

Evaluate

$$\frac{x^{3}}{3}+x^{2}-3x$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}-3x times \frac{3}{3}.

$$\frac{x^{3}}{3}+\frac{3\left(x^{2}-3x\right)}{3}$$

Since \frac{x^{3}}{3} and \frac{3\left(x^{2}-3x\right)}{3} have the same denominator, add them by adding their numerators.

$$\frac{x^{3}+3\left(x^{2}-3x\right)}{3}$$

Do the multiplications in x^{3}+3\left(x^{2}-3x\right).

$$\frac{x^{3}+3x^{2}-9x}{3}$$

Factor

$$\frac{x\left(x-\frac{-3\sqrt{5}-3}{2}\right)\left(x-\frac{3\sqrt{5}-3}{2}\right)}{3}$$