Show that #x^8-y^8=(x-y)(x+y)(x^2+y^2)(x^4+y^4)#?

2 Answers
P dilip_k
Jun 1, 2018

#x^8-y^8#

#=(x^4)^2-(y^4)^2#

#=(x^4-y^4)(x^4+y^4)#

#=((x^2)^2-(y^2)^2)(x^4+y^4)#

#=(x^2-y^2)(x^2+y^2)(x^4+y^4)#

#=(x-y)(x+y)(x^2+y^2)(x^4+y^4)#.

Jim G.
Jun 1, 2018

#"see explanation"#

Explanation:

#x^8-y^8" is a "color(blue)"difference of squares"#

#•color(white)(x)a^2-b^2=(a-b)(a+b)#

#"here "a=x^4" and "b=y^4#

#x^8-y^8=(x^4-y^4)(x^4+y^4)#

#x^4-y^4" is also a "color(blue)"difference of squares"#

#=(x^2-y^2)(x^2+y^2)(x^4+y^4)#

#x^2-y^2" is also a "color(blue)"difference of squares"#

#=(x-y)(x+y)(x^2+y^2)(x^4+y^4)larrcolor(red)" as required"#

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