how do you use the binomial theorem to find the value of \[{{999}^{3}}\]?

Hint: The binomial theorem describes the algebraic expansion of powers of a binomial. It is possible to expand the polynomial \[{{\left( a+b \right)}^{n}}\] into a sum involving terms of the form \[a{{x}^{b}}{{y}^{c}}\], where the exponents b and c are nonnegative integers with \[b+c=n\], and the coefficient a of each term is a specific positive integer depending on n and b. According to the theorem, it is possible to expand the polynomial
\[{{\left( a+b \right)}^{n}}\]in the form \[{{\left( a+b \right)}^{n}}=\sum\limits_{k=0}^{n}{\left( \begin{matrix}
   n \\
   k \\
\end{matrix} \right){{a}^{n-k}}{{b}^{k}}}\] where \[\left( \begin{matrix}
   n \\
   k \\
\end{matrix} \right)=\dfrac{n!}{k!\left( n-k \right)!}\].

Complete step by step answer:
As per the given question, we have to find the value of an expression using the binomial theorem. Here, we have to find the value of \[{{999}^{3}}\] using a binomial theorem.
We can write \[999=1000-1={{10}^{3}}-1\]. On substituting this value in \[{{999}^{3}}\] we can rewrite it as
\[\Rightarrow {{999}^{3}}={{\left( {{10}^{3}}-1 \right)}^{3}}\]
Now compare \[{{\left( {{10}^{3}}-1 \right)}^{3}}\] with \[{{\left( a+b \right)}^{n}}\].
On comparing we get \[a={{10}^{3}},b=-1,n=3\]. Now we expand the polynomial using the binomial theorem.
\[\Rightarrow \]\[{{\left( {{10}^{3}}-1 \right)}^{3}}=\sum\limits_{k=0}^{3}{\left( \begin{matrix}
   3 \\
   k \\
\end{matrix} \right){{\left( {{10}^{3}} \right)}^{3-k}}{{\left( -1 \right)}^{k}}}\]
Now we expand the summation. Then we get,
\[\Rightarrow \]\[{{\left( {{10}^{3}}-1 \right)}^{3}}=\left( \begin{matrix}
   3 \\
   0 \\
\end{matrix} \right){{\left( {{10}^{3}} \right)}^{3-0}}{{\left( -1 \right)}^{0}}+\left( \begin{matrix}
   3 \\
   1 \\
\end{matrix} \right){{\left( {{10}^{3}} \right)}^{3-1}}{{\left( -1 \right)}^{1}}+\left( \begin{matrix}
   3 \\
   2 \\
\end{matrix} \right){{\left( {{10}^{3}} \right)}^{3-2}}{{\left( -1 \right)}^{2}}+\left( \begin{matrix}
   3 \\
   3 \\
\end{matrix} \right){{\left( {{10}^{3}} \right)}^{3-3}}{{\left( -1 \right)}^{3}}\]
\[\Rightarrow \]\[{{\left( {{10}^{3}}-1 \right)}^{3}}=\left( \begin{matrix}
   3 \\
   0 \\
\end{matrix} \right){{\left( {{10}^{3}} \right)}^{3}}\left( 1 \right)+\left( \begin{matrix}
   3 \\
   1 \\
\end{matrix} \right){{\left( {{10}^{3}} \right)}^{2}}\left( -1 \right)+\left( \begin{matrix}
   3 \\
   2 \\
\end{matrix} \right){{\left( {{10}^{3}} \right)}^{1}}\left( 1 \right)+\left( \begin{matrix}
   3 \\
   3 \\
\end{matrix} \right){{\left( {{10}^{3}} \right)}^{0}}\left( -1 \right)\]
\[\Rightarrow \]\[{{\left( {{10}^{3}}-1 \right)}^{3}}=\left( \begin{matrix}
   3 \\
   0 \\
\end{matrix} \right)\left( {{10}^{9}} \right)\left( 1 \right)+\left( \begin{matrix}
   3 \\
   1 \\
\end{matrix} \right)\left( {{10}^{6}} \right)\left( -1 \right)+\left( \begin{matrix}
   3 \\
   2 \\
\end{matrix} \right)\left( {{10}^{3}} \right)\left( 1 \right)+\left( \begin{matrix}
   3 \\
   3 \\
\end{matrix} \right)\left( 1 \right)\left( -1 \right)\]
\[\Rightarrow \]\[{{\left( {{10}^{3}}-1 \right)}^{3}}=\left( \begin{matrix}
   3 \\
   0 \\
\end{matrix} \right)\left( {{10}^{9}} \right)-\left( \begin{matrix}
   3 \\
   1 \\
\end{matrix} \right)\left( {{10}^{6}} \right)+\left( \begin{matrix}
   3 \\
   2 \\
\end{matrix} \right)\left( {{10}^{3}} \right)-\left( \begin{matrix}
   3 \\
   3 \\
\end{matrix} \right)\left( 1 \right)\]
 We know that \[\left( \begin{matrix}
   n \\
   k \\
\end{matrix} \right)=\dfrac{n!}{k!\left( n-k \right)!}\] . From this the above values become \[\left( \begin{matrix}
   3 \\
   0 \\
\end{matrix} \right)=1,\left( \begin{matrix}
   3 \\
   1 \\
\end{matrix} \right)=3,\left( \begin{matrix}
   3 \\
   2 \\
\end{matrix} \right)=3,\] and \[\left( \begin{matrix}
   3 \\
   3 \\
\end{matrix} \right)=1\]. On substituting these values in the equation. The equation becomes
\[\Rightarrow {{\left( {{10}^{3}}-1 \right)}^{3}}=\left( 1 \right)\left( {{10}^{9}} \right)-\left( 3 \right)\left( {{10}^{6}} \right)+\left( 3 \right)\left( {{10}^{3}} \right)-\left( 1 \right)\left( 1 \right)\]
\[\Rightarrow {{\left( {{10}^{3}}-1 \right)}^{3}}=1000000000-3000000+3000-1\]
\[\Rightarrow {{\left( {{10}^{3}}-1 \right)}^{3}}=997002999\]
Therefore, the value of the given value \[{{999}^{3}}\] is 997002999.

Note:
 In order to solve such types of questions, we need to have enough knowledge of the binomial theorem and its expansions. In the above way, we can find expansions of many functions and also find the values of complex polynomial functions easily. While calculating \[\left( \begin{matrix}
   n \\
   k \\
\end{matrix} \right)\] calculate the factorials correctly without any mistakes. The binomial theorem helps us to find the value of the polynomials easily. We must avoid calculation mistakes to get the correct result or solution.