Question

Expand the binomial ${{\left( x-3 \right)}^{5}}$

Answer 1

Hint: Use binomial theorem for the expansion of the binomial ${{\left( x-3 \right)}^{5}}$. According to the binomial theorem, the expansion of the binomial ${{\left( x+y \right)}^{n}}$ is given by ${{\left( x+y \right)}^{n}}{{=}^{n}}{{C}_{0}}{{x}^{n}}{{+}^{n}}{{C}_{1}}{{x}^{n-1}}y+\cdots {{+}^{n}}{{C}_{n}}{{y}^{n}}$
Put y = -3 and n = 5 and use the fact that $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ to get the expansion of the term ${{\left( x-3 \right)}^{5}}$.

Complete step-by-step answer:
Alternatively, construct Pascal’s triangle till n=5 and hence find the expansion.
Alternatively, write ${{\left( x-3 \right)}^{5}}$ as ${{\left( x-3 \right)}^{3}}{{\left( x-3 \right)}^{2}}$. Use the fact that ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ and hence find the expansion of ${{\left( x-3 \right)}^{5}}$

We know that ${{\left( x+y \right)}^{n}}{{=}^{n}}{{C}_{0}}{{x}^{n}}{{+}^{n}}{{C}_{1}}{{x}^{n-1}}y+\cdots {{+}^{n}}{{C}_{n}}{{x}^{n}}$
Put y = -3 and n = 5, we get
${{\left( x-3 \right)}^{5}}{{=}^{5}}{{C}_{0}}{{x}^{5}}{{+}^{5}}{{C}_{1}}{{x}^{4}}\left( -3 \right){{+}^{5}}{{C}_{2}}{{x}^{3}}{{\left( -3 \right)}^{2}}{{+}^{5}}{{C}_{3}}{{x}^{2}}{{\left( -3 \right)}^{3}}{{+}^{5}}{{C}_{4}}x{{\left( -3 \right)}^{4}}{{+}^{5}}{{C}_{5}}{{\left( -3 \right)}^{5}}$
Now, we know that
$^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
Hence, we have
$^{5}{{C}_{0}}=\dfrac{5!}{5!0!}=1$
We know that
$\dfrac{^{n}{{C}_{r}}}{^{n}{{C}_{r-1}}}=\dfrac{n-r+1}{r}$
Put n = 5, r = 1, we get
$\begin{align}
  & \dfrac{^{5}{{C}_{1}}}{^{5}{{C}_{0}}}=\dfrac{5-1+1}{1}=5 \\
 & {{\Rightarrow }^{5}}{{C}_{1}}=5{{\times }^{5}}{{C}_{0}}=5 \\
\end{align}$
Put n =5, r = 2, we get
$\begin{align}
  & \dfrac{^{5}{{C}_{2}}}{^{5}{{C}_{1}}}=\dfrac{5-2+1}{2}=2 \\
 & {{\Rightarrow }^{5}}{{C}_{2}}=2{{\times }^{5}}{{C}_{1}}=2\times 5=10 \\
\end{align}$
Put n =5, r = 3, we get
$\begin{align}
  & \dfrac{^{5}{{C}_{3}}}{^{5}{{C}_{2}}}=\dfrac{5-3+1}{3}=1 \\
 & {{\Rightarrow }^{5}}{{C}_{3}}=1{{\times }^{5}}{{C}_{2}}=1\times 10=10 \\
\end{align}$
Put n =5, r = 4, we get
$\begin{align}
  & \dfrac{^{5}{{C}_{4}}}{^{5}{{C}_{3}}}=\dfrac{5-4+1}{4}=\dfrac{1}{2} \\
 & {{\Rightarrow }^{5}}{{C}_{4}}=\dfrac{1}{2}{{\times }^{5}}{{C}_{3}}=\dfrac{1}{2}\times 10=5 \\
\end{align}$
Put n =5, r = 5, we get
$\begin{align}
  & \dfrac{^{5}{{C}_{5}}}{^{5}{{C}_{4}}}=\dfrac{5-5+1}{5}=\dfrac{1}{5} \\
 & {{\Rightarrow }^{5}}{{C}_{5}}=\dfrac{1}{5}{{\times }^{5}}{{C}_{4}}=\dfrac{1}{5}\times 5=1 \\
\end{align}$
Hence, we have
${{\left( x-3 \right)}^{5}}={{x}^{5}}-3\times 5{{x}^{4}}+{{3}^{2}}\times 10{{x}^{3}}-{{3}^{3}}\times 10{{x}^{2}}+{{3}^{4}}\times 5x-{{3}^{5}}$
Simplifying, we get
${{\left( x-3 \right)}^{5}}={{x}^{5}}-15{{x}^{4}}+90{{x}^{3}}-270{{x}^{2}}+405x-243$, which is the required expansion of the given binomial.

Note: Alternative Solution:
Using Pascal’s triangle
\[\begin{matrix}
   {} & {} & {} & {} & {} & {} & 1 & {} & {} & {} & {} & {} & {} \\
   {} & {} & {} & {} & {} & 1 & {} & 1 & {} & {} & {} & {} & {} \\
   {} & {} & {} & {} & 1 & {} & 2 & {} & 1 & {} & {} & {} & {} \\
   {} & {} & {} & 1 & {} & 3 & {} & 3 & {} & 1 & {} & {} & {} \\
   {} & {} & 1 & {} & 4 & {} & 6 & {} & 4 & {} & 1 & {} & {} \\
   {} & 1 & {} & 5 & {} & 10 & {} & 10 & {} & 5 & {} & 1 & {} \\
\end{matrix}\]
Hence from pascal’s triangle, we have
${{\left( x-3 \right)}^{5}}={{x}^{5}}-3\times 5{{x}^{4}}+{{3}^{2}}\times 10{{x}^{3}}-{{3}^{3}}\times 10{{x}^{2}}+{{3}^{4}}\times 5x-{{3}^{5}}$
Simplifying, we get
${{\left( x-3 \right)}^{5}}={{x}^{5}}-15{{x}^{4}}+90{{x}^{3}}-270{{x}^{2}}+405x-243$, which is the required expansion of the given binomial.
Alternative Solution:
We know that ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$
Hence, we have
${{\left( x-3 \right)}^{3}}={{x}^{3}}-9{{x}^{2}}+27x-27$
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Hence, we have
${{\left( x-3 \right)}^{2}}={{x}^{2}}-6x+9$
Hence, we have
${{\left( x-3 \right)}^{5}}={{\left( x-3 \right)}^{2}}{{\left( x-3 \right)}^{3}}=\left( {{x}^{2}}-6x+9 \right)\left( {{x}^{3}}-9{{x}^{2}}+27x-27 \right)$
Expanding, we get
$\begin{align}
  & {{\left( x-3 \right)}^{5}}= \\
 & \begin{matrix}
   {{x}^{5}} & -9{{x}^{4}} & +27{{x}^{3}} & -27{{x}^{2}} & {} & {} \\
   {} & -6{{x}^{4}} & +54{{x}^{3}} & -162{{x}^{2}} & +162 & {} \\
   {} & {} & +9{{x}^{3}} & -81{{x}^{2}} & 243x & -243 \\
\end{matrix} \\
 & ={{x}^{5}}-15{{x}^{4}}+90{{x}^{3}}-270{{x}^{2}}+405x-243 \\
\end{align}$