Expand the expression ${\left( {2x - 3} \right)^6}$

Hint: In this question, we are given a binomial and we have to expand it using the binomial theorem. Here the power of the binomial is 6. The formula for binomial theorem is given by
$ {\left( {a + b} \right)^n} = \sum\limits_{r = 0}^n {\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right){a^{n - r}}{b^r}} $, where $\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right) = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Here, we have $a$ equal to $2x$ and $b$ equal to $3$ and $n$ equal to $6$. Using this formula we can expand the given binomial.

Complete step by step solution:
In this question, we are given a binomial in terms of x with power 6 and we need to expand it.
The given binomial is: ${\left( {2x - 3} \right)^6}$
To expand this binomial, we are going to use the binomial theorem.
So, let us see what binomial theorem is.
Binomial theorem is used to expand an expression raised to any positive finite power. Let us take an expression ${\left( {a + b} \right)^n}$. So, it can be expanded using binomial theorem as
$ {\left( {a + b} \right)^n} = \sum\limits_{r = 0}^n {\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right){a^{n - r}}{b^r}} $, where $\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right) = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Here, our binomial is ${\left( {2x - 3} \right)^6}$. So, here $a = 2x$ and $b = 3$.
So therefore, expansion of ${\left( {2x - 3} \right)^6}$ is
$
   \Rightarrow {\left( {2x - 3} \right)^6} = \sum\limits_{r = 0}^6 {\left( {\begin{array}{*{20}{c}}
  6 \\
  r
\end{array}} \right)} {\left( {2x} \right)^{6 - r}}{\left( 3 \right)^r} \\
   \Rightarrow {\left( {2x - 3} \right)^6} = \left( {\begin{array}{*{20}{c}}
  6 \\
  0
\end{array}} \right){\left( {2x} \right)^{6 - 0}}{\left( 3 \right)^0} - \left( {\begin{array}{*{20}{c}}
  6 \\
  1
\end{array}} \right){\left( {2x} \right)^{6 - 1}}{\left( 3 \right)^1} + \left( {\begin{array}{*{20}{c}}
  6 \\
  2
\end{array}} \right){\left( {2x} \right)^{6 - 2}}{\left( 3 \right)^2} - \left( {\begin{array}{*{20}{c}}
  6 \\
  3
\end{array}} \right){\left( {2x} \right)^{6 - 3}}{\left( 3 \right)^3} + \left( {\begin{array}{*{20}{c}}
  6 \\
  4
\end{array}} \right){\left( {2x} \right)^{6 - 4}}{\left( 3 \right)^4} - \left( {\begin{array}{*{20}{c}}
  6 \\
  5
\end{array}} \right){\left( {2x} \right)^{6 - 5}}{\left( 3 \right)^5} \\
   + \left( {\begin{array}{*{20}{c}}
  6 \\
  6
\end{array}} \right){\left( {2x} \right)^{6 - 6}}{\left( 3 \right)^6} \\
   \Rightarrow {\left( {2x - 3} \right)^6} = 1{\left( {2x} \right)^6}\left( 1 \right) - 6{\left( {2x} \right)^5}\left( 3 \right) + 15{\left( {2x} \right)^4}\left( 9 \right) - 20{\left( {2x} \right)^3}\left( {27} \right) + 15{\left( {2x} \right)^2}\left( {81} \right) - 6\left( {2x} \right)\left( {243} \right) + \\
  1{\left( {2x} \right)^0}\left( {729} \right) \\
   \Rightarrow {\left( {2x - 3} \right)^6} = 64{x^6} - 576{x^5} + 2160{x^4} - 4320{x^3} + 4860{x^2} - 2916x + 729 \\
$
Hence, we have found the expansion of ${\left( {2x - 3} \right)^6}$ using the binomial theorem. Therefore, the expansion of ${\left( {2x - 3} \right)^6}$ is $64{x^6} - 576{x^5} + 2160{x^4} - 4320{x^3} + 4860{x^2} - 2916x + 729$.

Note:
 Here, note that the given binomial is of the form ${\left( {a - b} \right)^n}$. So, we need to change the signs in the given formula. We have to use minus sign for every even term, that is for second term, fourth term, sixth term and so, we will use the minus sign. For example: if we have $n = 10$, then the signs will be as
$ + \left( {1st} \right) - \left( {2nd} \right) + \left( {3rd} \right) - \left( {4th} \right) + \left( {5th} \right) - \left( {6th} \right) + \left( {7th} \right) - \left( {8th} \right) + \left( {9th} \right) - \left( {10th} \right) + \left( {11th} \right)$.