Question

x⁸-y⁸ factorise please do fast it is urgent tomorrow my exam is going on​

Answer 1

Answer:

x8-y8

Final result :

(x4 + y4) • (x2 + y2) • (x + y) • (x - y)

Step by step solution :

Step 1 :

Trying to factor as a Difference of Squares :

1.1 Factoring: x8-y8

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =

A2 - AB + BA - B2 =

A2 - AB + AB - B2 =

A2 - B2

Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : x8 is the square of x4

Check : y8 is the square of y4

Factorization is : (x4 + y4) • (x4 - y4)

Trying to factor as a Difference of Squares :

1.2 Factoring: x4 - y4

Check : x4 is the square of x2

Check : y4 is the square of y2

Factorization is : (x2 + y2) • (x2 - y2)

Trying to factor as a Difference of Squares :

1.3 Factoring: x2 - y2

Check : x2 is the square of x1

Check : y2 is the square of y1

Factorization is : (x + y) • (x - y)

Final result :

(x4 + y4) • (x2 + y2) • (x + y) • (x - y)

Step-by-step explanation:

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Answer 2

$\bold\red{\underline{\underline{Answer:}}}$

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

$\bold\orange{Given:}$

$\bold{=>x^{8}-y^{8}}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{=>x^{8}-y^{8}}$

$\bold{=>By \ identity}$

$\bold{a^{2}-b^{2}=(a+b)(a-b)}$

$\bold{=>(x^{4}+y^{4})(x^{4}-y^{4})}$

$\bold{=>(x^{4}+y^{4})[(x^{2}+y^{2})(x^{2}-y^{2})]}$

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

$\bold\purple{\tt{\therefore{x^{8}-y^{8}=}}}$

$\bold\purple{\tt{(x^{4}+y^{4})(x^{2}+y^{2})(x+y)(x-y).}}$