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Question: How do you expand \( {{\left( x-1 \right)}^{3}} \) using binomial expression?...

How do you expand (x−1)3{{\left( x-1 \right)}^{3}} using binomial expression?

Explanation

Solution

Hint : We use the binomial form of n-degree of the difference of two terms (a−b)n=nC0anb0−nC1an−1b1+nC2an−2b2−....+(−1)rnCran−rbr+....+(−1)nnCna0bn{{\left( a-b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}-{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}-....+{{\left( -1 \right)}^{r}}{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+....+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}} . We put the values n=3,a=x,b=1n=3,a=x,b=1 to get the simplified form of (x−1)3{{\left( x-1 \right)}^{3}} . We can also find the simplification of the given polynomial (x−1)3{{\left( x-1 \right)}^{3}} according to the identity (a−b)3=a3−3a2b+3ab2−b3{{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}} .

Complete step-by-step answer :
We can use the binomial theorem to find the general form and then put the value of 3.
The binomial form of n-degree polynomial of subtraction of two numbers can be expressed as (a−b)n=nC0anb0−nC1an−1b1+nC2an−2b2−....+(−1)rnCran−rbr+....+(−1)nnCna0bn{{\left( a-b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}-{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}-....+{{\left( -1 \right)}^{r}}{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+....+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}} .
We need to find the cube of difference of two numbers. So, we put n=3n=3. We get
(a−b)3=3C0a3b0−3C1a3−1b1+3C2a3−2b2−3C3a3−3b3=a3−3a2b+3ab2−b3{{\left( a-b \right)}^{3}}={}^{3}{{C}_{0}}{{a}^{3}}{{b}^{0}}-{}^{3}{{C}_{1}}{{a}^{3-1}}{{b}^{1}}+{}^{3}{{C}_{2}}{{a}^{3-2}}{{b}^{2}}-{}^{3}{{C}_{3}}{{a}^{3-3}}{{b}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}} .
Then we directly put the values of a=x,b=1a=x,b=1 to find the simplification
(x−1)3=3C0x310−3C1x3−111+3C2x3−212−3C3x3−313=x3−3x2+3x−1{{\left( x-1 \right)}^{3}}={}^{3}{{C}_{0}}{{x}^{3}}{{1}^{0}}-{}^{3}{{C}_{1}}{{x}^{3-1}}{{1}^{1}}+{}^{3}{{C}_{2}}{{x}^{3-2}}{{1}^{2}}-{}^{3}{{C}_{3}}{{x}^{3-3}}{{1}^{3}}={{x}^{3}}-3{{x}^{2}}+3x-1 .
We can also write the simplification in the form of
(x−1)3=x3−3x2+3x−1=x3−3x(x−1)−1{{\left( x-1 \right)}^{3}}={{x}^{3}}-3{{x}^{2}}+3x-1={{x}^{3}}-3x\left( x-1 \right)-1 .
So, the correct answer is “Option B”.

Note : We need to find the simplified form of (x−1)3{{\left( x-1 \right)}^{3}} . This is the cube of difference of two numbers. We know that (a−b)2=a2+b2−2ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab .
We need to multiply the term (a−b)\left( a-b \right) on both side of the identity (a−b)2=a2+b2−2ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab .
On the left side of the equation, we get (a−b)2(a−b)=(a−b)3{{\left( a-b \right)}^{2}}\left( a-b \right)={{\left( a-b \right)}^{3}} .
On the right side we have (a2+b2−2ab)(a−b)\left( {{a}^{2}}+{{b}^{2}}-2ab \right)\left( a-b \right) . We use multiplication and get
⇒(a2+b2−2ab)(a−b) =a2.a+a.b2−2ab×a−a2.b−b2.b+2ab.b =a3+ab2−2a2b−a2b−b3+2ab2 =a3−3a2b+3ab2−b3  \Rightarrow \left( {{a}^{2}}+{{b}^{2}}-2ab \right)\left( a-b \right) \\\ ={{a}^{2}}.a+a.{{b}^{2}}-2ab\times a-{{a}^{2}}.b-{{b}^{2}}.b+2ab.b \\\ ={{a}^{3}}+a{{b}^{2}}-2{{a}^{2}}b-{{a}^{2}}b-{{b}^{3}}+2a{{b}^{2}} \\\ ={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}} \\\
We also can take another form where
(a−b)3=a3−3a2b+3ab2−b3=a3−b3−3ab(a−b){{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right) .
Now we replace the values for a=x,b=1a=x,b=1 in the equation (a−b)3=a3−3a2b+3ab2−b3{{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}} .
(x−1)3 =x3−3x2×1+3x×12−13 =x3−3x2+3x−1  {{\left( x-1 \right)}^{3}} \\\ ={{x}^{3}}-3{{x}^{2}}\times 1+3x\times {{1}^{2}}-{{1}^{3}} \\\ ={{x}^{3}}-3{{x}^{2}}+3x-1 \\\