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Question: Expand the binomial \({{\left( x-3 \right)}^{5}}\)...

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

Explanation

Solution

Hint: Use binomial theorem for the expansion of the binomial (x−3)5{{\left( x-3 \right)}^{5}}. According to the binomial theorem, the expansion of the binomial (x+y)n{{\left( x+y \right)}^{n}} is given by (x+y)n=nC0xn+nC1xn−1y+⋯+nCnyn{{\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 nCr=n!(n−r)!r!^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!} to get the expansion of the term (x−3)5{{\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 (x−3)5{{\left( x-3 \right)}^{5}} as (x−3)3(x−3)2{{\left( x-3 \right)}^{3}}{{\left( x-3 \right)}^{2}}. Use the fact that (a−b)3=a3−b3−3a2b+3ab2{{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}} and (a−b)2=a2−2ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} and hence find the expansion of (x−3)5{{\left( x-3 \right)}^{5}}

We know that (x+y)n=nC0xn+nC1xn−1y+⋯+nCnxn{{\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
(x−3)5=5C0x5+5C1x4(−3)+5C2x3(−3)2+5C3x2(−3)3+5C4x(−3)4+5C5(−3)5{{\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
nCr=n!(n−r)!r!^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}
Hence, we have
5C0=5!5!0!=1^{5}{{C}_{0}}=\dfrac{5!}{5!0!}=1
We know that
nCrnCr−1=n−r+1r\dfrac{^{n}{{C}_{r}}}{^{n}{{C}_{r-1}}}=\dfrac{n-r+1}{r}
Put n = 5, r = 1, we get
5C15C0=5−1+11=5 ⇒5C1=5×5C0=5 \begin{aligned} & \dfrac{^{5}{{C}_{1}}}{^{5}{{C}_{0}}}=\dfrac{5-1+1}{1}=5 \\\ & {{\Rightarrow }^{5}}{{C}_{1}}=5{{\times }^{5}}{{C}_{0}}=5 \\\ \end{aligned}
Put n =5, r = 2, we get
5C25C1=5−2+12=2 ⇒5C2=2×5C1=2×5=10 \begin{aligned} & \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{aligned}
Put n =5, r = 3, we get
5C35C2=5−3+13=1 ⇒5C3=1×5C2=1×10=10 \begin{aligned} & \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{aligned}
Put n =5, r = 4, we get
5C45C3=5−4+14=12 ⇒5C4=12×5C3=12×10=5 \begin{aligned} & \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{aligned}
Put n =5, r = 5, we get
5C55C4=5−5+15=15 ⇒5C5=15×5C4=15×5=1 \begin{aligned} & \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{aligned}
Hence, we have
(x−3)5=x5−3×5x4+32×10x3−33×10x2+34×5x−35{{\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
(x−3)5=x5−15x4+90x3−270x2+405x−243{{\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

{} & {} & {} & {} & {} & {} & 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 $\begin{aligned} & {{\left( x-3 \right)}^{5}}= \\\ & \begin{matrix} {{x}^{5}} & -9{{x}^{4}} & +27{{x}^{3}} & -27{{x}^{2}} & {} & {} \\\ {} & -6{{x}^{4}} & +54{{x}^{3}} & -162{{x}^{2}} & +162 & {} \\\ {} & {} & +9{{x}^{3}} & -81{{x}^{2}} & 243x & -243 \\\ \end{matrix} \\\ & ={{x}^{5}}-15{{x}^{4}}+90{{x}^{3}}-270{{x}^{2}}+405x-243 \\\ \end{aligned}$