Solve x^6-x^3=

$$x ^ { 6 } - x ^ { 3 } =$$

Factor

$$\left(x-1\right)\left(x^{2}+x+1\right)x^{3}$$

Solution Steps

Factor out x^{3}.

$$x^{3}\left(x^{3}-1\right)$$

Consider x^{3}-1. Rewrite x^{3}-1 as x^{3}-1^{3}. The difference of cubes can be factored using the rule: a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right).

$$\left(x-1\right)\left(x^{2}+x+1\right)$$

Rewrite the complete factored expression. Polynomial x^{2}+x+1 is not factored since it does not have any rational roots.

$$x^{3}\left(x-1\right)\left(x^{2}+x+1\right)$$

Evaluate

$$x^{3}\left(x^{3}-1\right)$$