How do you expand ${{\left( 3x+2y \right)}^{4}}$ ?
Answer
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Hint: We can expand the given expression ${{\left( 3x+2y \right)}^{4}}$ by first taking the square of $\left( 3x+2y \right)$ and then take the square of the result of the square of $\left( 3x+2y \right)$. We will take the square of $\left( 3x+2y \right)$ by using the algebraic identity which is equal to ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$. Then when you take the second time square of $\left( 3x+2y \right)$, you will need the algebraic identity which is equal to: ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right)$.
Complete step by step answer:
The algebraic expression given in the above problem is as follows:
${{\left( 3x+2y \right)}^{4}}$
We can rewrite the above expression as follows:
${{\left( {{\left( 3x+2y \right)}^{2}} \right)}^{2}}$
Now, taking the square of $\left( 3x+2y \right)$ using the following algebraic identity i.e.:
${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
When using the above algebraic identity, substituting $''a=3x''$ and $''b=2y''$ in the above identity we get,
$\begin{align}
& {{\left( 3x+2y \right)}^{2}}={{\left( 3x \right)}^{2}}+2\left( 3x \right)\left( 2y \right)+{{\left( 2y \right)}^{2}} \\
& \Rightarrow {{\left( 3x+2y \right)}^{2}}=9{{x}^{2}}+12xy+4{{y}^{2}} \\
\end{align}$
Now, taking square on both the sides we get,
${{\left( {{\left( 3x+2y \right)}^{2}} \right)}^{2}}={{\left( 9{{x}^{2}}+12xy+4{{y}^{2}} \right)}^{2}}$ ………. Eq. (1)
Applying the following algebraic identity on R.H.S of the above equation we get,
${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right)$
Substituting $a=9{{x}^{2}},b=12xy,c=4{{y}^{2}}$ in the above equation we get,
${{\left( 9{{x}^{2}}+12xy+4{{y}^{2}} \right)}^{2}}={{\left( 9{{x}^{2}} \right)}^{2}}+{{\left( 12xy \right)}^{2}}+{{\left( 4{{y}^{2}} \right)}^{2}}+2\left( \left( 9{{x}^{2}} \right)\left( 12xy \right)+\left( 12xy \right)\left( 4{{y}^{2}} \right)+\left( 4{{y}^{2}} \right)\left( 9{{x}^{2}} \right) \right)$
Multiplying 2 with the terms written in the bracket we get,
$\begin{align}
& \Rightarrow {{\left( 9{{x}^{2}}+12xy+4{{y}^{2}} \right)}^{2}}=81{{x}^{4}}+144{{x}^{2}}{{y}^{2}}+16{{y}^{4}}+2\left( 108{{x}^{3}}y+48x{{y}^{3}}+36{{x}^{2}}{{y}^{2}} \right) \\
& \Rightarrow {{\left( 9{{x}^{2}}+12xy+4{{y}^{2}} \right)}^{2}}=81{{x}^{4}}+144{{x}^{2}}{{y}^{2}}+16{{y}^{4}}+216{{x}^{3}}y+96x{{y}^{3}}+72{{x}^{2}}{{y}^{2}} \\
\end{align}$
Adding the coefficients of ${{x}^{2}}{{y}^{2}}$ in the R.H.S of the above equation we get,
$\begin{align}
& {{\left( 9{{x}^{2}}+12xy+4{{y}^{2}} \right)}^{2}}=81{{x}^{4}}+\left( 144+72 \right){{x}^{2}}{{y}^{2}}+16{{y}^{4}}+216{{x}^{3}}y+96x{{y}^{3}} \\
& \Rightarrow {{\left( 9{{x}^{2}}+12xy+4{{y}^{2}} \right)}^{2}}=81{{x}^{4}}+216{{x}^{2}}{{y}^{2}}+16{{y}^{4}}+216{{x}^{3}}y+96x{{y}^{3}} \\
\end{align}$
Substituting the above value in eq. (1) we get,
${{\left( {{\left( 3x+2y \right)}^{2}} \right)}^{2}}=$$81{{x}^{4}}+216{{x}^{2}}{{y}^{2}}+16{{y}^{4}}+216{{x}^{3}}y+96x{{y}^{3}}$
Hence, the expansion of ${{\left( 3x+2y \right)}^{4}}$ is equal to:
$81{{x}^{4}}+216{{x}^{2}}{{y}^{2}}+16{{y}^{4}}+216{{x}^{3}}y+96x{{y}^{3}}$
Note: The alternate approach to the above problem is as follows:
We know that the binomial expansion of:
${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+......+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$
So, using the above expansion in expanding ${{\left( 3x+2y \right)}^{4}}$ we get,
${{\left( 3x+2y \right)}^{4}}={}^{4}{{C}_{0}}{{\left( 3x \right)}^{4}}{{\left( 2y \right)}^{0}}+{}^{4}{{C}_{1}}{{\left( 3x \right)}^{3}}{{\left( 2y \right)}^{1}}+{}^{4}{{C}_{2}}{{\left( 3x \right)}^{2}}{{\left( 2y \right)}^{2}}+{}^{4}{{C}_{3}}{{\left( 3x \right)}^{1}}{{\left( 2y \right)}^{3}}+{}^{4}{{C}_{4}}{{\left( 3x \right)}^{0}}{{\left( 2y \right)}^{4}}$ We know that: ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ so using this expansion in the above equation we get,
$\begin{align}
& {{\left( 3x+2y \right)}^{4}}=\dfrac{4!}{0!\left( 4-0 \right)!}{{\left( 3x \right)}^{4}}{{\left( 2y \right)}^{0}}+\dfrac{4!}{1!\left( 4-1 \right)!}{{\left( 3x \right)}^{3}}{{\left( 2y \right)}^{1}}+\dfrac{4!}{2!\left( 4-2 \right)!}{{\left( 3x \right)}^{2}}{{\left( 2y \right)}^{2}}+\dfrac{4!}{3!\left( 4-3 \right)!}{{\left( 3x \right)}^{1}}{{\left( 2y \right)}^{3}}+\dfrac{4!}{4!\left( 4-4 \right)!}{{\left( 3x \right)}^{0}}{{\left( 2y \right)}^{4}} \\
& \Rightarrow {{\left( 3x+2y \right)}^{4}}=\dfrac{4!}{0!\left( 4 \right)!}{{\left( 3x \right)}^{4}}{{\left( 2y \right)}^{0}}+\dfrac{4!}{1!\left( 3 \right)!}{{\left( 3x \right)}^{3}}{{\left( 2y \right)}^{1}}+\dfrac{4!}{2!\left( 2 \right)!}{{\left( 3x \right)}^{2}}{{\left( 2y \right)}^{2}}+\dfrac{4!}{3!\left( 1 \right)!}{{\left( 3x \right)}^{1}}{{\left( 2y \right)}^{3}}+\dfrac{4!}{4!\left( 0 \right)!}{{\left( 3x \right)}^{0}}{{\left( 2y \right)}^{4}} \\
\end{align}$
We also know that the value of $0!=1$ so using this relation in the above we get,
$\begin{align}
& {{\left( 3x+2y \right)}^{4}}=\dfrac{4!}{\left( 1 \right)\left( 4 \right)!}{{\left( 3x \right)}^{4}}{{\left( 2y \right)}^{0}}+\dfrac{4!}{1!\left( 3 \right)!}{{\left( 3x \right)}^{3}}{{\left( 2y \right)}^{1}}+\dfrac{4!}{2!\left( 2 \right)!}{{\left( 3x \right)}^{2}}{{\left( 2y \right)}^{2}}+\dfrac{4!}{3!\left( 1 \right)!}{{\left( 3x \right)}^{1}}{{\left( 2y \right)}^{3}}+\dfrac{4!}{4!\left( 1 \right)}{{\left( 3x \right)}^{0}}{{\left( 2y \right)}^{4}} \\
& \Rightarrow {{\left( 3x+2y \right)}^{4}}=\dfrac{1}{\left( 1 \right)1}{{\left( 3x \right)}^{4}}{{\left( 2y \right)}^{0}}+\dfrac{4.3!}{1!\left( 3 \right)!}{{\left( 3x \right)}^{3}}{{\left( 2y \right)}^{1}}+\dfrac{4.3.2!}{2!\left( 2 \right)!}{{\left( 3x \right)}^{2}}{{\left( 2y \right)}^{2}}+\dfrac{4.3!}{3!\left( 1 \right)!}{{\left( 3x \right)}^{1}}{{\left( 2y \right)}^{3}}+\dfrac{4!}{4!\left( 1 \right)}{{\left( 3x \right)}^{0}}{{\left( 2y \right)}^{4}} \\
\end{align}$
In the above, $3!,4!,2!$ will be cancelled out from the numerator and the denominator in the R.H.S of the above equation we get,
$\begin{align}
& {{\left( 3x+2y \right)}^{4}}=81{{x}^{4}}+\dfrac{4}{1}54{{x}^{3}}y+\dfrac{4.3}{2.1}36{{x}^{2}}{{y}^{2}}+\dfrac{4}{\left( 1 \right)}24x{{y}^{3}}+\dfrac{1}{\left( 1 \right)}16{{y}^{4}} \\
& \Rightarrow {{\left( 3x+2y \right)}^{4}}=81{{x}^{4}}+216{{x}^{3}}y+216{{x}^{2}}{{y}^{2}}+96x{{y}^{3}}+16{{y}^{4}} \\
\end{align}$
Complete step by step answer:
The algebraic expression given in the above problem is as follows:
${{\left( 3x+2y \right)}^{4}}$
We can rewrite the above expression as follows:
${{\left( {{\left( 3x+2y \right)}^{2}} \right)}^{2}}$
Now, taking the square of $\left( 3x+2y \right)$ using the following algebraic identity i.e.:
${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
When using the above algebraic identity, substituting $''a=3x''$ and $''b=2y''$ in the above identity we get,
$\begin{align}
& {{\left( 3x+2y \right)}^{2}}={{\left( 3x \right)}^{2}}+2\left( 3x \right)\left( 2y \right)+{{\left( 2y \right)}^{2}} \\
& \Rightarrow {{\left( 3x+2y \right)}^{2}}=9{{x}^{2}}+12xy+4{{y}^{2}} \\
\end{align}$
Now, taking square on both the sides we get,
${{\left( {{\left( 3x+2y \right)}^{2}} \right)}^{2}}={{\left( 9{{x}^{2}}+12xy+4{{y}^{2}} \right)}^{2}}$ ………. Eq. (1)
Applying the following algebraic identity on R.H.S of the above equation we get,
${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right)$
Substituting $a=9{{x}^{2}},b=12xy,c=4{{y}^{2}}$ in the above equation we get,
${{\left( 9{{x}^{2}}+12xy+4{{y}^{2}} \right)}^{2}}={{\left( 9{{x}^{2}} \right)}^{2}}+{{\left( 12xy \right)}^{2}}+{{\left( 4{{y}^{2}} \right)}^{2}}+2\left( \left( 9{{x}^{2}} \right)\left( 12xy \right)+\left( 12xy \right)\left( 4{{y}^{2}} \right)+\left( 4{{y}^{2}} \right)\left( 9{{x}^{2}} \right) \right)$
Multiplying 2 with the terms written in the bracket we get,
$\begin{align}
& \Rightarrow {{\left( 9{{x}^{2}}+12xy+4{{y}^{2}} \right)}^{2}}=81{{x}^{4}}+144{{x}^{2}}{{y}^{2}}+16{{y}^{4}}+2\left( 108{{x}^{3}}y+48x{{y}^{3}}+36{{x}^{2}}{{y}^{2}} \right) \\
& \Rightarrow {{\left( 9{{x}^{2}}+12xy+4{{y}^{2}} \right)}^{2}}=81{{x}^{4}}+144{{x}^{2}}{{y}^{2}}+16{{y}^{4}}+216{{x}^{3}}y+96x{{y}^{3}}+72{{x}^{2}}{{y}^{2}} \\
\end{align}$
Adding the coefficients of ${{x}^{2}}{{y}^{2}}$ in the R.H.S of the above equation we get,
$\begin{align}
& {{\left( 9{{x}^{2}}+12xy+4{{y}^{2}} \right)}^{2}}=81{{x}^{4}}+\left( 144+72 \right){{x}^{2}}{{y}^{2}}+16{{y}^{4}}+216{{x}^{3}}y+96x{{y}^{3}} \\
& \Rightarrow {{\left( 9{{x}^{2}}+12xy+4{{y}^{2}} \right)}^{2}}=81{{x}^{4}}+216{{x}^{2}}{{y}^{2}}+16{{y}^{4}}+216{{x}^{3}}y+96x{{y}^{3}} \\
\end{align}$
Substituting the above value in eq. (1) we get,
${{\left( {{\left( 3x+2y \right)}^{2}} \right)}^{2}}=$$81{{x}^{4}}+216{{x}^{2}}{{y}^{2}}+16{{y}^{4}}+216{{x}^{3}}y+96x{{y}^{3}}$
Hence, the expansion of ${{\left( 3x+2y \right)}^{4}}$ is equal to:
$81{{x}^{4}}+216{{x}^{2}}{{y}^{2}}+16{{y}^{4}}+216{{x}^{3}}y+96x{{y}^{3}}$
Note: The alternate approach to the above problem is as follows:
We know that the binomial expansion of:
${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+......+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$
So, using the above expansion in expanding ${{\left( 3x+2y \right)}^{4}}$ we get,
${{\left( 3x+2y \right)}^{4}}={}^{4}{{C}_{0}}{{\left( 3x \right)}^{4}}{{\left( 2y \right)}^{0}}+{}^{4}{{C}_{1}}{{\left( 3x \right)}^{3}}{{\left( 2y \right)}^{1}}+{}^{4}{{C}_{2}}{{\left( 3x \right)}^{2}}{{\left( 2y \right)}^{2}}+{}^{4}{{C}_{3}}{{\left( 3x \right)}^{1}}{{\left( 2y \right)}^{3}}+{}^{4}{{C}_{4}}{{\left( 3x \right)}^{0}}{{\left( 2y \right)}^{4}}$ We know that: ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ so using this expansion in the above equation we get,
$\begin{align}
& {{\left( 3x+2y \right)}^{4}}=\dfrac{4!}{0!\left( 4-0 \right)!}{{\left( 3x \right)}^{4}}{{\left( 2y \right)}^{0}}+\dfrac{4!}{1!\left( 4-1 \right)!}{{\left( 3x \right)}^{3}}{{\left( 2y \right)}^{1}}+\dfrac{4!}{2!\left( 4-2 \right)!}{{\left( 3x \right)}^{2}}{{\left( 2y \right)}^{2}}+\dfrac{4!}{3!\left( 4-3 \right)!}{{\left( 3x \right)}^{1}}{{\left( 2y \right)}^{3}}+\dfrac{4!}{4!\left( 4-4 \right)!}{{\left( 3x \right)}^{0}}{{\left( 2y \right)}^{4}} \\
& \Rightarrow {{\left( 3x+2y \right)}^{4}}=\dfrac{4!}{0!\left( 4 \right)!}{{\left( 3x \right)}^{4}}{{\left( 2y \right)}^{0}}+\dfrac{4!}{1!\left( 3 \right)!}{{\left( 3x \right)}^{3}}{{\left( 2y \right)}^{1}}+\dfrac{4!}{2!\left( 2 \right)!}{{\left( 3x \right)}^{2}}{{\left( 2y \right)}^{2}}+\dfrac{4!}{3!\left( 1 \right)!}{{\left( 3x \right)}^{1}}{{\left( 2y \right)}^{3}}+\dfrac{4!}{4!\left( 0 \right)!}{{\left( 3x \right)}^{0}}{{\left( 2y \right)}^{4}} \\
\end{align}$
We also know that the value of $0!=1$ so using this relation in the above we get,
$\begin{align}
& {{\left( 3x+2y \right)}^{4}}=\dfrac{4!}{\left( 1 \right)\left( 4 \right)!}{{\left( 3x \right)}^{4}}{{\left( 2y \right)}^{0}}+\dfrac{4!}{1!\left( 3 \right)!}{{\left( 3x \right)}^{3}}{{\left( 2y \right)}^{1}}+\dfrac{4!}{2!\left( 2 \right)!}{{\left( 3x \right)}^{2}}{{\left( 2y \right)}^{2}}+\dfrac{4!}{3!\left( 1 \right)!}{{\left( 3x \right)}^{1}}{{\left( 2y \right)}^{3}}+\dfrac{4!}{4!\left( 1 \right)}{{\left( 3x \right)}^{0}}{{\left( 2y \right)}^{4}} \\
& \Rightarrow {{\left( 3x+2y \right)}^{4}}=\dfrac{1}{\left( 1 \right)1}{{\left( 3x \right)}^{4}}{{\left( 2y \right)}^{0}}+\dfrac{4.3!}{1!\left( 3 \right)!}{{\left( 3x \right)}^{3}}{{\left( 2y \right)}^{1}}+\dfrac{4.3.2!}{2!\left( 2 \right)!}{{\left( 3x \right)}^{2}}{{\left( 2y \right)}^{2}}+\dfrac{4.3!}{3!\left( 1 \right)!}{{\left( 3x \right)}^{1}}{{\left( 2y \right)}^{3}}+\dfrac{4!}{4!\left( 1 \right)}{{\left( 3x \right)}^{0}}{{\left( 2y \right)}^{4}} \\
\end{align}$
In the above, $3!,4!,2!$ will be cancelled out from the numerator and the denominator in the R.H.S of the above equation we get,
$\begin{align}
& {{\left( 3x+2y \right)}^{4}}=81{{x}^{4}}+\dfrac{4}{1}54{{x}^{3}}y+\dfrac{4.3}{2.1}36{{x}^{2}}{{y}^{2}}+\dfrac{4}{\left( 1 \right)}24x{{y}^{3}}+\dfrac{1}{\left( 1 \right)}16{{y}^{4}} \\
& \Rightarrow {{\left( 3x+2y \right)}^{4}}=81{{x}^{4}}+216{{x}^{3}}y+216{{x}^{2}}{{y}^{2}}+96x{{y}^{3}}+16{{y}^{4}} \\
\end{align}$
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