help please

x⁸-y⁸ 

difference of cubes

(x²-y²)(x⁴+x²y²+y⁴) 

first polynomial can be factored further

(x+y)(x-y)    (x⁴+x²y²+y⁴)

next polynomial can be factored further

x⁴ + 2x²y²-x²y²+y⁴

x⁴+2x²y²+y⁴- x²y²

^{first three terms factorable by perfect square}

(x²+y²)² - x²y²

(x²+y²)² - (xy)²

difference of squares again

(x²+y²+xy)(x²+y²-xy)

first polynomials factored form and second polynomials factored form put together

(x+y)(x-y)(x²+y²+xy)(x²+y²-xy)

sorry if formatting is weird

Completely factor x^8 - y^8.

\(\begin{array}{|rcll|} \hline \mathbf{x^8 - y^8} &=& \left( x^\frac{8}{2} - y^\frac{8}{2} \right)\left( x^\frac{8}{2} + y^\frac{8}{2} \right) \\ &=& \left( x^4 - y^4 \right)\left( x^4 + y^4 \right) \\ &=& \left( x^\frac{4}{2} - y^\frac{4}{2} \right)\left( x^\frac{4}{2} + y^\frac{4}{2} \right)\left( x^4 + y^4 \right) \\ &=& \left( x^2 - y^2 \right)\left( x^2 + y^2 \right) \left( x^4 + y^4 \right)\\ &=& \left( x^\frac{2}{2} - y^\frac{2}{2} \right)\left( x^\frac{2}{2} + y^\frac{2}{2} \right) \left( x^2 + y^2 \right) \left( x^4 + y^4 \right) \\ \mathbf{x^8 - y^8} &=&\mathbf{ (x - y)(x + y)\left( x^2 + y^2 \right) \left( x^4 + y^4 \right) } \\ \hline \end{array} \)