Question

x12-y12 factorise the polynomial

Answer 1

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Answer 2

Answer:

The factorization is

$x^{12}-y^{12}=(x-y)(x+y)(x^2 + xy +y^2)(x^2-xy+y^2)(x^6+y^6)$

Step-by-step explanation:

Given the polynomial

$x^{12}-y^{12}$

we have to factorize the above polynomial.

$x^{12}-y^{12}$

$(x^6)^2-(y^6)^2$

By identity: $a^2-b^2=(a-b)(a+b)$

$Put, a=x^6, b=y^6$

$(x^6)^2-(y^6)^2$

$(x^6-y^6)(x^6+y^6)$

$((x^3)^2-(y^3)^2)(x^6+y^6)$

$(x^3-y^3)(x^3+y^3)(x^6+y^6)$

By identity,

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

$a^3 - b^3 = (a – b)(a^2 + ab + b^2)$

Hence, the factorization becomes

$(x-y)(x+y)(x^2 + xy +y^2)(x^2-xy+y^2)(x^6+y^6)$

which is required factored form.