Answer
The expansion of ${{\left( {{x}^{3}}+2y \right)}^{5}}={{x}^{15}}+10{{x}^{12}}y+40{{x}^{9}}{{y}^{2}}+80{{x}^{6}}{{y}^{3}}+80{{x}^{3}}{{y}^{4}}+32{{y}^{5}}$.
Work Step by Step
We know that the general formula of a binomial expansion is:
${{\left( a+b \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{a}^{n-r}}{{b}^{r}}$
Where, ${}^{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$
Then, expand ${{\left( {{x}^{3}}+2y \right)}^{5}}$ in this form as:
$\begin{align}
& {{\left( {{x}^{3}}+2y \right)}^{5}}=\sum\limits_{r=0}^{5}{{}^{5}{{C}_{r}}}{{\left( {{x}^{3}} \right)}^{5-r}}{{\left( 2y \right)}^{r}} \\
& =\left[ {}^{5}{{C}_{0}}{{\left( {{x}^{3}} \right)}^{5-0}}{{\left( 2y \right)}^{0}}+{}^{5}{{C}_{1}}{{\left( {{x}^{3}} \right)}^{5-1}}{{\left( 2y \right)}^{1}}+{}^{5}{{C}_{2}}{{\left( {{x}^{3}} \right)}^{5-2}}{{\left( 2y \right)}^{2}}+{}^{5}{{C}_{3}}{{\left( {{x}^{3}} \right)}^{5-3}}{{\left( 2y \right)}^{3}}+{}^{5}{{C}_{4}}{{\left( {{x}^{3}} \right)}^{5-4}}{{\left( 2y \right)}^{4}}+{}^{5}{{C}_{5}}{{\left( {{x}^{3}} \right)}^{5-5}}{{\left( 2y \right)}^{5}} \right] \\
& ={{x}^{15}}+10{{x}^{12}}y+40{{x}^{9}}{{y}^{2}}+80{{x}^{6}}{{y}^{3}}+80{{x}^{3}}{{y}^{4}}+32{{y}^{5}}
\end{align}$
Thus, the expansion is ${{x}^{15}}+10{{x}^{12}}y+40{{x}^{9}}{{y}^{2}}+80{{x}^{6}}{{y}^{3}}+80{{x}^{3}}{{y}^{4}}+32{{y}^{5}}$.
