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0
51
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Completely factor x^8 - y^8.

Dec 3, 2019

#1
+15
0

x8-y

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

x+ 2x2y2-x2y2+y4

x4+2x2y2+y4- x2y2

first three terms factorable by perfect square

(x2+y2)2 - x2y2

(x2+y2)- (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

Dec 3, 2019
#2
+23872
+1

Completely factor x^8 - y^8.

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

Dec 3, 2019