Question

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

Answer 1

Hint: We use the binomial form of n-degree of the difference of two terms $ {{\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=1 $ to get the simplified form of $ {{\left( x-1 \right)}^{3}} $ . We can also find the simplification of the given polynomial $ {{\left( x-1 \right)}^{3}} $ according to the identity $ {{\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 $ {{\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=3 $. We get
 $ {{\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=1 $ to find the simplification
 $ {{\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
 $ {{\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 $ {{\left( x-1 \right)}^{3}} $ . This is the cube of difference of two numbers. We know that $ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ .
We need to multiply the term $ \left( a-b \right) $ on both side of the identity $ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ .
On the left side of the equation, we get $ {{\left( a-b \right)}^{2}}\left( a-b \right)={{\left( a-b \right)}^{3}} $ .
On the right side we have $ \left( {{a}^{2}}+{{b}^{2}}-2ab \right)\left( a-b \right) $ . We use multiplication and get
 $
   \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
 $ {{\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=1 $ in the equation $ {{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}} $ .
 $
   {{\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 \\
$