Question

Factorise: $x^6+7x^3-8$

Answer 1

Let $x^3=a$, then the equation $x^6+7x^3-8$ becomes $a^2+7a-8$

Consider the equation $a^2+7a-8$ and factorise it as follows:

$a^2+7a-8=a^2+8a-a-8=a(a+8)-1(a+8)=(a-1)(a+8)$

Now, substitute the value of $a$ as $a=x^3$:

$a^2+7a-8=a^2+8a-a-8=a(a+8)-1(a+8)=(a-1)(a+8)=(x^3-1)(x^3+8)=(x^3-1^3)(x^3+2^3)$

$=(x-1)(x^2+x+1)(x+2)(x^2-2x+4)$

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

Hence, $x^6+7x^3-8=(x-1)(x^2+x+1)(x+2)(x^2-2x+4)$