Solve the following expression:
\[64{{a}^{3}}-27{{b}^{3}}-144{{a}^{2}}b+108a{{b}^{2}}\]
(a) \[{{\left( 4a-9b \right)}^{3}}\]
(b) \[{{\left( 2a+3b \right)}^{3}}\]
(c) \[{{\left( 4a-3b \right)}^{3}}\]
(d) \[{{\left( a+9b \right)}^{3}}\]

Hint: In this question, we first need to write the given equation in terms of like terms. Then by using the factorisation of polynomials formula we can further sove it and write in the simplified form.
\[{{\left( x-a \right)}^{3}}={{x}^{3}}-3{{x}^{2}}a+3{{a}^{2}}x-{{a}^{3}}\]

Complete step-by-step answer:
Now, from the given expression in the question we have
\[\Rightarrow 64{{a}^{3}}-27{{b}^{3}}-144{{a}^{2}}b+108a{{b}^{2}}\]
Polynomial:
An expression of the form \[{{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}+....+{{a}_{n-1}}x+{{a}_{n}}\], where $a_0$, $a_1$,....., $a_n$ are real numbers and n is a non-negative integer, is called a polynomial.
Cubic Polynomial: A polynomial of degree three is called cubic polynomial.
As we already know that from the factorisation of the polynomials we have
\[{{\left( x-a \right)}^{3}}={{x}^{3}}-3{{x}^{2}}a+3{{a}^{2}}x-{{a}^{3}}\]
Let us now consider the given expression in the question
\[\Rightarrow 64{{a}^{3}}-27{{b}^{3}}-144{{a}^{2}}b+108a{{b}^{2}}\]
Now, this can also be written as
\[\Rightarrow 64{{a}^{3}}-144{{a}^{2}}b+108a{{b}^{2}}-27{{b}^{3}}\]
Let us now rewrite the above expression to simplify it further
\[\Rightarrow 64{{a}^{3}}-48\times 3\times {{a}^{2}}b+27\times 4\times a{{b}^{2}}-27{{b}^{3}}\]
Now, this can be further written in the expanded form as
\[\Rightarrow {{\left( 4a \right)}^{3}}-3\times 16\times 3\times {{a}^{2}}b+3\times 9\times 4\times a{{b}^{2}}-{{\left( 3b \right)}^{3}}\]
Now, from the above factorisation of polynomials formula we have
\[{{\left( x-a \right)}^{3}}={{x}^{3}}-3{{x}^{2}}a+3{{a}^{2}}x-{{a}^{3}}\]
Now, by using the factorisation of polynomials formula the above expression can be further written as
\[\Rightarrow {{\left( 4a \right)}^{3}}-3\times {{\left( 4 \right)}^{2}}\times 3\times {{a}^{2}}b+3\times {{\left( 3 \right)}^{2}}\times 4\times a{{b}^{2}}-{{\left( 3b \right)}^{3}}\]
Now, on comparing this expression with the above formula we get,
\[\begin{align}
  & x=4a \\
 & a=3b \\
\end{align}\]
Now, by using the formula of factorisation of polynomials we can further write it in the simplified form as
\[\Rightarrow {{\left( 4a-3b \right)}^{3}}\]
Hence, the correct option is (c).

Note: Instead of considering the formula of \[{{\left( x-a \right)}^{3}}\] from the factorisation of polynomials we can also use the formula of \[{{\left( x+a \right)}^{3}}\] in which the sign of a should also be considered according to the given expression. Both the methods give the same result.
It is important to note that while comparing the given expression with the formula we have we first need to convert the given expression accordingly such that it can be expressed in terms of the expanded form of the formula we have.