Solve x^12-3/3=

$$x ^ { 12 } - \frac { 3 } { 3 } =$$

Factor

$$\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)\left(x^{4}-x^{2}+1\right)$$

Solution Steps

Rewrite x^{12}-1 as \left(x^{6}\right)^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Consider x^{6}-1. Rewrite x^{6}-1 as \left(x^{3}\right)^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

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).

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

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

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

Consider x^{6}+1. Rewrite x^{6}+1 as \left(x^{2}\right)^{3}+1^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

$$\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: x^{2}-x+1,x^{2}+x+1,x^{4}-x^{2}+1,x^{2}+1.

$$\left(x-1\right)\left(x^{2}-x+1\right)\left(x+1\right)\left(x^{2}+x+1\right)\left(x^{4}-x^{2}+1\right)\left(x^{2}+1\right)$$

Evaluate

$$\left(x^{4}-1\right)\left(x^{4}-x^{2}+1\right)\left(\left(x^{2}+1\right)^{2}-x^{2}\right)$$

Solution Steps

Divide 3 by 3 to get 1.

$$x^{12}-1$$