Use the Binomial Theorem to expand the binomial, (3x+y)^3 and express the result in simplified form

Given that:
\(\displaystyle{\left({3}{x}+{y}\right)}^{{3}}\)
The standard form of binomial theorem for any positive integer “n” is given by,
\(\displaystyle{\left({a}+{b}\right)}^{{n}}={C}_{{0}}{a}^{{n}}+{C}_{{1}}{a}^{{{n}-{1}}}{b}+{C}_{{2}}{a}^{{{n}-{2}}}{b}^{{2}}+\ldots+{C}_{{{n}-{1}}}{a}\cdot{b}^{{{n}-{1}}}+{C}_{{n}}{b}^{{n}}\)
Here, a=3x, b=y, n=3
To expand the given polynomials
\(\displaystyle{\left({3}{x}+{y}\right)}^{{3}}={C}_{{0}}{\left({3}{x}\right)}^{{3}}+{C}_{{1}}{\left({3}{x}\right)}^{{{3}-{1}}}{y}+{C}_{{2}}{\left({3}{x}\right)}^{{{3}-{2}}}{y}^{{2}}+{C}_{{3}}{\left({3}{x}\right)}^{{{3}-{3}}}{y}^{{3}}\)
\(\displaystyle={\frac{{{3}!}}{{{0}!{\left({3}-{0}\right)}!}}}{\left({27}{x}^{{3}}\right)}+{\frac{{{3}!}}{{{1}!{\left({3}-{1}\right)}!}}}{\left({3}{x}\right)}^{{2}}{y}+{\frac{{{3}!}}{{{2}!{\left({3}-{2}\right)}!}}}{\left({3}{x}\right)}^{{1}}{y}^{{2}}+{\frac{{{3}!}}{{{3}!{\left({3}-{3}\right)}!}}}{\left({3}{x}\right)}^{{0}}{y}^{{3}}\)
\(\displaystyle{\frac{{{3}!}}{{{1}\cdot{3}!}}}{\left({27}{x}^{{3}}\right)}+{\frac{{{3}\cdot{2}\cdot{1}}}{{{1}{\left({2}\times{1}\right)}}}}\)\(9x^2y+\frac{3\cdot2\cdot1}{(2\times1)\cdot1}3xy^2+\frac{3\cdot2\cdot1}{3\cdot2\cdot1(1)}y^3\)
\(\displaystyle={1}\cdot{\left({27}{x}^{{3}}\right)}+{3}{\left({9}{x}^{{2}}{y}\right)}+{3}{\left({3}{x}{y}^{{2}}\right)}+{y}^{{3}}\)
\(\displaystyle={27}{x}^{{3}}+{27}{x}^{{2}}{y}+{9}{x}{y}^{{2}}+{y}^{{3}}\)