Factorise the cubic polynomial $54{{x}^{3}}-16{{y}^{3}}$?

Hint: We can observe that the given equation contains two cubic variables in subtraction. In algebra we have the formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$. But in the given equation we have additionally coefficients for the variables, so we need to factorize the coefficients of the given variables and we will convert the given equation in form of ${{a}^{3}}-{{b}^{3}}$, by applying some exponential rules, then we will use the algebraic formula and write the given equation as the product of its factors.

Complete step-by-step solution:
Given equation $54{{x}^{3}}-16{{y}^{3}}$.
Considering the coefficient of ${{x}^{3}}$, which is $54$. Factorizing $54$.
Dividing $54$ with $2$, then we will get zero reminder and $27$ as quotient. So, we can write $54$ as
$54=2\times 27$.
Now we will factorize $27$. We know that $27$ is not divisible by $2$. So, checking whether the $27$ is divisible by $3$ or not. $27$ is divisible by $3$ and we will get $9$ as the quotient, then we can write $27$ as
$27=3\times 9$
Now we will factorize $9$. We know that we can write $9$ as $9=3\times 3$. From all the above values, the value of $54$ can be
$\begin{align}
  & 54=2\times 27 \\
 & \Rightarrow 54=2\times 3\times 9 \\
 & \Rightarrow 54=2\times 3\times 3\times 3 \\
\end{align}$
We have an exponential rule $a\times a\times a\times ....n\text{ times}={{a}^{n}}$, then we will have
$\therefore 54=2\times {{3}^{3}}$
Considering the coefficient of ${{y}^{3}}$, which is $16$. Factoring $16$.
Dividing $16$ with $2$, then we will get zero remainder and $8$ as quotient. So, we can write $16$ as
$16=2\times 8$.
Now we will factorize $8$. We know that $8$ is divisible by $2$and it gives $4$ quotient, then we can write $8$ as
$8=2\times 4$
Now we will factorize $4$. We know that we can write $4$ as $4=2\times 2$. From all the above values, the value of $16$ can be
$\begin{align}
  & 16=2\times 8 \\
 & \Rightarrow 16=2\times 2\times 4 \\
 & \Rightarrow 16=2\times 2\times 2\times 2 \\
\end{align}$
We have an exponential rule $a\times a\times a\times ....n\text{ times}={{a}^{n}}$, then we will have
$\therefore 16={{2}^{4}}$.
Now the given equation is modified as
$54{{x}^{3}}-16{{y}^{3}}=\left( 2\times {{3}^{3}} \right){{x}^{3}}-\left( {{2}^{4}} \right){{y}^{3}}$
Taking $2$ as common in the left part
$\Rightarrow 54{{x}^{3}}-16{{y}^{3}}=2\left[ {{3}^{3}}{{x}^{3}}-{{2}^{3}}{{y}^{3}} \right]$
We have the exponential rule ${{a}^{m}}{{b}^{m}}={{\left( ab \right)}^{m}}$, then we will get
$\Rightarrow 54{{x}^{3}}-16{{y}^{3}}=2\left[ {{\left( 3x \right)}^{3}}-{{\left( 2y \right)}^{3}} \right]$
Applying the formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ in the above equation, then we will get
$\begin{align}
  & \Rightarrow 54{{x}^{3}}-16{{y}^{3}}=2\left[ \left( 3x-2y \right)\left( {{\left( 3x \right)}^{2}}+\left( 3x \right)\left( 2y \right)+{{\left( 2y \right)}^{2}} \right) \right] \\
 & \Rightarrow 54{{x}^{3}}-16{{y}^{3}}=2\left( 3x-2y \right)\left( 9{{x}^{2}}+6xy+4{{y}^{2}} \right) \\
\end{align}$

Note: We can check whether the obtained answer is correct or not by multiplying the factors we have in the final equation with each other and equate it to the given equation. If we get both the equations are equal then the answer is correct, otherwise we did a mistake in the solution.