How do you expand the given binomial ${{\left( 2x+4 \right)}^{3}}$?

Hint: We start solving the problem by recalling the binomial expansion of ${{\left( a+b \right)}^{n}}$ as ${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+...+{}^{n}{{C}_{n}}{{b}^{n}}$. We use this expansion for the given binomial and then make the necessary calculations to proceed through the problem. We then make use of the results that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$ and $0!=1$ in the obtained result to proceed further through the problem. We then make the necessary calculations to get the required expansion of the given binomial.

Complete step by step answer:
According to the problem, we are asked to expand the given binomial ${{\left( 2x+4 \right)}^{3}}$.
We have given the binomial ${{\left( 2x+4 \right)}^{3}}$ ---(1).
From the binomial expansion of ${{\left( a+b \right)}^{n}}$, we know that ${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+...+{}^{n}{{C}_{n}}{{b}^{n}}$. Let us use this result in equation (1).
$\Rightarrow {{\left( 2x+4 \right)}^{3}}={}^{3}{{C}_{0}}{{\left( 2x \right)}^{3}}+{}^{3}{{C}_{1}}{{\left( 2x \right)}^{3-1}}\left( 4 \right)+{}^{3}{{C}_{2}}{{\left( 2x \right)}^{3-2}}{{\left( 4 \right)}^{2}}+{}^{3}{{C}_{3}}{{\left( 4 \right)}^{3}}$.
$\Rightarrow {{\left( 2x+4 \right)}^{3}}={}^{3}{{C}_{0}}{{\left( 2x \right)}^{3}}+{}^{3}{{C}_{1}}{{\left( 2x \right)}^{2}}\left( 4 \right)+{}^{3}{{C}_{2}}{{\left( 2x \right)}^{1}}{{\left( 4 \right)}^{2}}+{}^{3}{{C}_{3}}{{\left( 4 \right)}^{3}}$.
$\Rightarrow {{\left( 2x+4 \right)}^{3}}={}^{3}{{C}_{0}}\left( 8{{x}^{3}} \right)+{}^{3}{{C}_{1}}\left( 4{{x}^{2}} \right)\left( 4 \right)+{}^{3}{{C}_{2}}\left( 2x \right)\left( 16 \right)+{}^{3}{{C}_{3}}\left( 64 \right)$.
$\Rightarrow {{\left( 2x+4 \right)}^{3}}={}^{3}{{C}_{0}}\left( 8{{x}^{3}} \right)+{}^{3}{{C}_{1}}\left( 16{{x}^{2}} \right)+{}^{3}{{C}_{2}}\left( 32x \right)+{}^{3}{{C}_{3}}\left( 64 \right)$ ---(2).
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$ and $0!=1$. Let us use these results in equation (2).
$\Rightarrow {{\left( 2x+4 \right)}^{3}}=\left( \dfrac{3!}{0!3!} \right)\left( 8{{x}^{3}} \right)+\left( \dfrac{3!}{1!2!} \right)\left( 16{{x}^{2}} \right)+\left( \dfrac{3!}{2!1!} \right)\left( 32x \right)+\left( \dfrac{3!}{3!0!} \right)\left( 64 \right)$.
$\Rightarrow {{\left( 2x+4 \right)}^{3}}=\left( 1 \right)\left( 8{{x}^{3}} \right)+\left( 3 \right)\left( 16{{x}^{2}} \right)+\left( 3 \right)\left( 32x \right)+\left( 1 \right)\left( 64 \right)$.
$\Rightarrow {{\left( 2x+4 \right)}^{3}}=8{{x}^{3}}+48{{x}^{2}}+96x+64$.
So, we have found the expansion of the given binomial ${{\left( 2x+4 \right)}^{3}}$ as $8{{x}^{3}}+48{{x}^{2}}+96x+64$.
$\therefore $ The required expansion of the given binomial ${{\left( 2x+4 \right)}^{3}}$ as $8{{x}^{3}}+48{{x}^{2}}+96x+64$.

Note:
Whenever we get this type of problem, we make use of the binomial expansion of ${{\left( a+b \right)}^{n}}$ to get the required answer. We should perform each step carefully in order to avoid confusion and calculation mistakes while solving this problem. We can also solve the given problem by making use of the pascal’s triangle which will also give the similar result. Similarly, we can expect problems to find the expansion of the given binomial ${{\left( 2x+\dfrac{3}{x} \right)}^{4}}$.