Question

factorise
x^8 - y^8
​

Answer 1

Answer:

HEy... Mate here's ur aNswer✌️

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

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Answer 2

Answer to your question is

$(x^4+y^4)(x^2+y^2)(x+y)(x-y)$

I'll answer briefly.

Step-by-step explanation:

To express (square)-(square) in multiplication form,

we're gonna be using identity which is $a^2-b^2=(a+b)(a-b)$

(factorization)

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

From that, we can observe that last bracket is able to be factorized.

(factorization)

$(x^4+y^4)(x^{2\times 2}-y^{2\times 2})=(x^4+y^4)(x^2+y^2)(x^2-y^2)$

Last bracket is still factorized.

(factorization)

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

Last bracket is unable to be factorized.

Therefore, factorization is ended.