Solve x^3/3-2x^2+3x=

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

Evaluate

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

Short Solution Steps

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

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

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

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

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

$$\frac{x^{3}-6x^{2}+9x}{3}$$

Factor

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

Solution Steps

Factor out \frac{1}{3}.

$$\frac{x^{3}-6x^{2}+9x}{3}$$

Consider x^{3}-6x^{2}+9x. Factor out x.

$$x\left(x^{2}-6x+9\right)$$

Consider x^{2}-6x+9. Use the perfect square formula, a^{2}-2ab+b^{2}=\left(a-b\right)^{2}, where a=x and b=3.

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

Rewrite the complete factored expression.

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