Solve X^9-1

$$X ^ { 9 } - 1$$

Factor

$$\left(X-1\right)\left(X^{2}+X+1\right)\left(X^{6}+X^{3}+1\right)$$

Solution Steps

Rewrite X^{9}-1 as \left(X^{3}\right)^{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).

$$\left(X^{3}-1\right)\left(X^{6}+X^{3}+1\right)$$

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. The following polynomials are not factored since they do not have any rational roots: X^{2}+X+1,X^{6}+X^{3}+1.

$$\left(X-1\right)\left(X^{2}+X+1\right)\left(X^{6}+X^{3}+1\right)$$

Evaluate

$$X^{9}-1$$