How do you factor x^3 - 1?

2 Answers
Rhys
Dec 15, 2017

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!

Jim G.
Dec 30, 2017

(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)

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