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

Hint: For answering this question we have been asked to factorize the given expression $8{{x}^{3}}-{{y}^{3}}$. We will use the formula for factorization given as ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$. If we observe the given expression is in the form of the left hand side one. By expressing it as $\Rightarrow {{\left( 2x \right)}^{3}}-{{\left( y \right)}^{3}}$ .

Complete step by step solution:
Now considering from the question we need to factorize the given expression $8{{x}^{3}}-{{y}^{3}}$ .
From the basic concepts we know that there exists a formula for factorization mathematically given as ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$
As we know that the number eight can be written as the cube of two. Hence the given expression can be simply written as $ 8{{x}^{3}}-{{y}^{3}}={{\left( 2x \right)}^{3}}-{{\left( y \right)}^{3}}$ .
By applying the formulae we will have $\Rightarrow {{\left( 2x \right)}^{3}}-{{\left( y \right)}^{3}}=\left( 2x-y \right)\left( 4{{x}^{2}}+{{y}^{2}}+2xy \right)$ .
Therefore we can conclude that the factors of the given expression $8{{x}^{3}}-{{y}^{3}}$ are$\left( 2x-y \right),\left( 4{{x}^{2}}+{{y}^{2}}+2xy \right)$.

Note: While answering this question we need to take care of our calculations we perform and concepts that we apply while solving. This is a very simple and time efficient question. Very few mistakes are possible here. Similarly we have another formula for factorization that is mathematically given as ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$ .
For example we can factorize the given expression by considering it as $ 8{{x}^{3}}+{{\left( -y \right)}^{3}}$ and apply then we will have $ {{\left( 2x \right)}^{3}}+{{\left( -y \right)}^{3}}=\left( 2x-y \right)\left( 4{{x}^{2}}+{{y}^{2}}+2xy \right)$. It is similar to the one we got above. The proof for this factorization expression can be obtained by using the formulae ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$ , ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$ ,${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ . The proof for ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$ goes on as
$\begin{align}
  & {{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right) \\
 & \Rightarrow {{\left( a-b \right)}^{3}}+3ab\left( a-b \right)={{a}^{3}}-{{b}^{3}} \\
 & \Rightarrow \left( a-b \right)\left[ {{\left( a-b \right)}^{2}}+3ab \right]={{a}^{3}}-{{b}^{3}} \\
 & \Rightarrow \left( a-b \right)\left[ {{a}^{2}}+{{b}^{2}}+ab \right]={{a}^{3}}-{{b}^{3}} \\
\end{align}$
Similarly we can prove the other formulae.