Solve 64x^6-y^6

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

Factor

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

Solution Steps

Rewrite 64x^{6}-y^{6} as \left(8x^{3}\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).

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

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

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

Consider 8x^{3}+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).

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

Rewrite the complete factored expression.

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

Evaluate

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