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

2 Answers
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!

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)#