How do you factor ##8x^3-8##?

Answer: ##8(x-1)(x^2+x+1)##

To factorize the expression, first we remove any obvious common factors of all the terms. Here, 8 divides every term, so we can extract the 8:

##8(x^3-1)##

Next we see if there are any factors of the expression ##(x^3-1)##. If there is a solution ##\alpha## to the equation ##x^3-1 = 0##, then ##(x-\alpha)## will divide ##(x^3 - 1)## (This is called the 'factor theorem'.) Here, ##x = 1## is an obvious solution, so we can take out ##(x-1)## as a factor,

##8(x-1)f(x)##

where ##(x-1)f(x) = (x^3 - 1)##, and ##f(x)## is a polynomial we must work out. ##f(x)## must be quadratic (degree 2), since ##x^3-1## has degree 3, so we can write it as a general quadratic:

##f(x) = ax^2 + bx + c##,

where we need to work out the values of a, b and c. Multiplying out the brackets:

##x^3 - 1 = (x-1)f(x) = (x-1)(ax^2 + bx + c)## ##= ax^3 + bx^2 + cx - ax^2 - bx - c## ##=ax^3 + (b-a)x^2 + (c-b)x -c##,

and by comparing the coefficients of the powers of x, we see that ##a=1##, ##b-a = 0##, ##c-b=0## and ##c=1##, implying that ##b=1##.

So we have our newly factored polynomial:

##8(x-1)(x^2 + x + 1)##

Can we factorize this further? In other words, are there any solutions to the equation ##x^2 + x + 1 = 0##? By using the quadratic formula, we find that the solutions are given by:

##x = frac(-1 + \sqrt(-3))(2)## or ##frac (-1 - \sqrt(-3))(2)##

Both of these solutions, however, involve taking the root of a negative number, which does not give a real number. Therefore the equation has been factorized as much as possible in the real numbers, and our job is complete:

##8x^3 - 8 = 8(x-1)(x^2 + x + 1) \quad\quad\quad\quad\quad\quad\square##