Solve x^12-y^6

$$x ^ { 12 } - y ^ { 6 }$$

Factor

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

Solution Steps

Rewrite x^{12}-y^{6} as \left(x^{6}\right)^{2}-\left(y^{3}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x^{6}-y^{3}\right)\left(x^{6}+y^{3}\right)$$

Consider x^{6}-y^{3}. Rewrite x^{6}-y^{3} as \left(x^{2}\right)^{3}-y^{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).

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

Consider x^{6}+y^{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).

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

Rewrite the complete factored expression.

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

Evaluate

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