Question

How do you factor `x^3 - 1`?

Answer 1

Expanding upon prior answer:

Explanation:

I want to expand upon an idea expressed in the prior answer

The idea of:

`(x^n - 1)/(x-1) = sum_(r=1) ^n x^(n-r)`

or not in sigma notation:

`(x^n -1 )/(x-1) = x^(n-1) + x^(n-2) + ... + x + 1`

We can prove this via induction:

Basis case :

`=> n = 1`

`LHS: (x^1-1)/(x-1) = 1`

`RHS: x^(1-1) = x^0 = 1`

Hence basis case holds

Induction:

Assume `n=k` holds:

`(x^k - 1)/(x-1) = sum_(r=1) ^k x^(k-r)`

`n = k+1`:

`sum_(r=1) ^(k+1) x^(k+1-r) = (sum_(r=1) ^k x^(k+1-r)) +1`

`= x *(sum_(r=1) ^(k) x^(k-r)) + 1`

`= x * ( (x^k -1)/(x-1) ) + 1`

`= (x^(k+1) - x)/(x-1) + 1`

`= (x^(k+1) - x) / (x-1) + (x-1)/(x-1)`

`= (x^(k+1) - 1 )/(x-1)`

Hence this is also what we yield when plugging directly into formula:

Hence holds for all `k in ZZ^+` and all `k+1 in ZZ^+` so holds for all `n in ZZ^+`

`=>` Proven by mathematical induction

I thought this was a nice idea to consider!

Answer 2

`(x-1)(x^2+x+1)`

Explanation:

`x^3-1" is a "color(blue)"difference of cubes"`

`•color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)`

`"here "a=x" and "b=1`

`rArrx^3-1=(x-1)(x^2+x+1)`