Question

How do you factorise the expression $27{x^3} - 8?$

Answer 1

Hint: As we know that factorising is the reverse of expanding brackets, it is an important way of solving equations. The first step of factoring an expression is to take out any common factors which the terms have. So if we were asked to factor the expression ${x^2} + x$, since $x$ goes into both terms, we would write $x(x + 1)$. Here we will use identities which will help us to factorise an algebraic expression easily i.e. ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$.

Complete step by step answer:
We will solve the above equation by applying the cubes formula, since both terms are perfect cubes.
As we know that the required formula is
${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$.
Here $27{x^3}$ and $8$, both are perfect cubes as we can rewrite $27{x^3}$ as ${(3x)^3}$ and 8 as ${2^3}$.
So $a = 3x$ and $b = 2$, now by applying formula the above given equation can be written as,
$ = (3x - 2)\{ {(3x)^2} + 3x \times 2 + {(2)^2}\} $
$ \Rightarrow (3x - 2)(9{x^2} + 6x + 4)$.

Hence the required answer is $(3x - 2)(9{x^2} + 6x + 4)$.

Note: We should keep in mind while solving these expressions that we use correct identities to factorise the given algebraic expressions and keep checking the negative and positive sign otherwise it will give the wrong answer. Also we should be careful while solving the difference of cubes identities and solve them accordingly. The above used formula of ${a^3} - {b^3}$ should not be confused with the other formula which is of ${(a - b)^3}$ as both of them are different. These are some of the standard algebraic identities. This is as far we can go with real coefficients as the remaining quadratic factors all have complex zeros.