Solve x^2-4x+4divx^3-8=

$${ x }^{ 2 } -4x+4 \div { x }^{ 3 } -8=$$

Evaluate

$$x^{2}-4x-8+\frac{4}{x^{3}}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}-4x-8 times \frac{x^{3}}{x^{3}}.

$$\frac{\left(x^{2}-4x-8\right)x^{3}}{x^{3}}+\frac{4}{x^{3}}$$

Since \frac{\left(x^{2}-4x-8\right)x^{3}}{x^{3}} and \frac{4}{x^{3}} have the same denominator, add them by adding their numerators.

$$\frac{\left(x^{2}-4x-8\right)x^{3}+4}{x^{3}}$$

Do the multiplications in \left(x^{2}-4x-8\right)x^{3}+4.

$$\frac{x^{5}-4x^{4}-8x^{3}+4}{x^{3}}$$

Factor

$$\frac{x^{5}-4x^{4}-8x^{3}+4}{x^{3}}$$