Question

factorise x^6-7x^3-8

Answer 1

Here is your answer

Hope it helps you

Answer 2

Answer:

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

Step-by-step explanation:

Given : Expression $x^6-7x^3-8$

To find : Factories the expression ?

Solution :

We can re-write the expression as

$x^6-7x^3-8=(x^3)^2-7(x^3)-8$

This is in the form of quadratic equation $ax^2+bx+c$

Applying middle term split,

$(x^3)^2-8(x^3)+x^3-8$

$(x^3)(x^3-8)+1(x^3-8)$

$(x^3+1)(x^3-8)$

Re-write as $(x^3+1^3)(x^3-2^3)$

Applying formula,

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

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

Therefore, The factories form of the expression is

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