Question

How do you factor: y = $27{{x}^{3}}-1$ ?

Answer 1

Hint: In this problem, we have to factorise the given expression using algebraic identities. We have to use the identity: ${{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)$. This is the cubic identity so cube root of 27 will be needed. Cube is of the form $x\times x\times x={{x}^{3}}$ and cube root is of the form $\sqrt[3]{{{x}^{3}}}=x$. So these are the main terms which are important to solve this question.

Complete step by step answer:
Now, let’s solve the question.
As we already know that algebraic identities are those which are used in factorisation of the polynomials. When we say polynomials, it means the expression which contains one or more than one term. They are of 3 types: monomial, binomial and trinomial. Monomial is the type which contains only one single term. Binomial one contains only two terms. And the trinomial contains 3 terms. The identities usually make the computations easy. There are various identities which need to be discussed here.
$\begin{align}
  & \Rightarrow {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
 & \Rightarrow {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \\
 & \Rightarrow {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) \\
 & \Rightarrow \left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab \\
 & \Rightarrow {{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right) \\
 & \Rightarrow {{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right) \\
 & \Rightarrow {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right) \\
 & \Rightarrow {{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) \\
\end{align}$
These are some of the basic identities which are used.
We are given:
$\Rightarrow $y = $27{{x}^{3}}-1$
By applying ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ we have to solve the given expression. Here a = 3x and b = 1:
$\Rightarrow $y = ${{\left( 3x \right)}^{3}}-{{1}^{3}}$
$\Rightarrow $y = $\left( 3x-1 \right)\left( {{\left( 3x \right)}^{2}}+3x\times 1+{{1}^{2}} \right)$
Solve further:
$\Rightarrow $y = $\left( 3x-1 \right)\left( 9{{x}^{2}}+3x+1 \right)$
So this is the factorisation of the given expression.

Note:
Take care of the closing brackets while squaring the numbers. Remove the brackets after squaring just like it is done in the last step. Use appropriate identities. There is a huge difference in ${{\left( a+b \right)}^{3}}$ and ${{a}^{3}}+{{b}^{3}}$. Both the identities are used differently.