Math, asked by sharmaartmamta373537, 1 year ago

factorise
x^8 - y^8

Answers

Answered by Anonymous
4

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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Answered by TakenName
0

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.

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