Question

How do you factor ${{x}^{3}}-8{{y}^{3}}$ ?

Answer 1

Hint: At first, we express $8$ as ${{2}^{3}}$ . Thereafter, we apply the formula of cubes which is ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ . We then check if the expression $\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ can be further factorised. If not, this will be the factorised form.

Complete step by step solution:
The given expression is,
${{x}^{3}}-8{{y}^{3}}$
We can see that the given expression contains only two terms. The algebraic part in both the terms have a cubic exponent. From this, we can get an intuition that we have to factorise the given expression in some cubic form. We see that $x$ and $y$ have power $3$ already. But, can we express $8$ in some cubic form? The answer is yes. $8$ is the cube of the natural number $2$ , that is, ${{2}^{3}}=8$ . We now rewrite the given expression as,
$\Rightarrow {{x}^{3}}-{{2}^{3}}{{y}^{3}}$
Now, we know the formula that ${{a}^{n}}{{b}^{n}}={{\left( ab \right)}^{n}}$ . Comparing it with the above expression, we see that in the above expression, $a=2,b=y,n=3$ . So, applying the formula, we get ${{2}^{3}}{{y}^{3}}={{\left( 2y \right)}^{3}}$ . Now, we rewrite the above expression as,
$\Rightarrow {{x}^{3}}-{{\left( 2y \right)}^{3}}$
Now, we know the formula of cubes. According to the formula, a cubic expression of the form ${{a}^{3}}-{{b}^{3}}$ can be written as $\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ . Comparing this formula with the above expression, we can see that $a=x,b=2y$ . So, we can rewrite the above expression as,
$\Rightarrow \left( x-2y \right)\left( {{x}^{2}}+x\left( 2y \right)+{{\left( 2y \right)}^{2}} \right)$
The above expression can be simplified as,
$\Rightarrow \left( x-2y \right)\left( {{x}^{2}}+2xy+4{{y}^{2}} \right)$
The expression in the second bracket is in its simplest form and cannot be further factorised or simplified. Therefore, we can conclude that the given expression can be factored to $\left( x-2y \right)\left( {{x}^{2}}+2xy+4{{y}^{2}} \right)$ .

Note: For these types for problems, we have no other option than to memorise some basic factorisation formulae. Only then can we solve them. We can also solve the problem in another way. We can see that if we put $x=2y$ in the given expression, we get $0$ . This means $x-2y$ is a factor. Dividing the expression by $x-2y$ , we get another factor. The product form of the two factors is the factorised form.