Solve 64x^6-y^12

$$64 x ^ { 6 } - y ^ { 12 }$$

Factor

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

Solution Steps

Rewrite 64x^{6}-y^{12} as \left(8x^{3}\right)^{2}-\left(y^{6}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(8x^{3}-y^{6}\right)\left(8x^{3}+y^{6}\right)$$

Consider 8x^{3}-y^{6}. Rewrite 8x^{3}-y^{6} as \left(2x\right)^{3}-\left(y^{2}\right)^{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).

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

Consider 8x^{3}+y^{6}. Rewrite 8x^{3}+y^{6} as \left(2x\right)^{3}+\left(y^{2}\right)^{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).

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

Rewrite the complete factored expression.

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

Evaluate

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