Question

Factor the expression by treating it as the difference of two squares, (x3)2-1

Answer 1

we are given

$(x^3)^2-1$

Let's assume x^3 as a

$(x^3)^2-1=a^2-1$

now, we can factor

and we get

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

now, we can plug back 'a'

and we get

$(x^3)^2-1=(x^3-1)(x^3+1)$

now, we can factor again

we know that

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

so, we get

$(x^3)^2-1=(x^3-1)(x+1)(x^2-x+1)$

now, we can use formula

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

we can plug it

$(x^3)^2-1=(x-1)(x^2+x+1)(x+1)(x^2-x+1)$

we can rearrange it

$(x^3)^2-1=(x-1)(x+1)(x^2-x+1)(x^2+x+1)$..............Answer