Question

How do you expand the binomial \[{{\left( x-y \right)}^{10}}\] ?

Answer 1

Hint: From the given question we are asked to expand the binomial\[{{\left( x-y \right)}^{10}}\]. To expand this, we have to use binomial theorem i.e., the expansion of \[{{\left( a+b \right)}^{n}}=\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}.\left( {{a}^{n-k}}{{b}^{k}} \right)}\]. Here we have to substitute x in place of a and \[\left( -y \right)\] in place of b. By using this formula in the binomial theorem we can expand the above binomial\[{{\left( x-y \right)}^{10}}\].

Complete step by step solution:
From the given question we have to expand the binomial \[{{\left( x-y \right)}^{10}}\]
As we know that we have to expand this by using binomial theorem. Binomial theorem describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the polynomial \[{{\left( a+b \right)}^{n}}\] into a sum involving terms of the form \[c{{a}^{x}}{{b}^{y}}\], where the exponents x and y are nonnegative integers with \[x+y=n\], and the coefficient c of each term is a specific positive integer depending on n and x. the coefficient c in the term of \[c{{a}^{x}}{{b}^{y}}\] is known as the binomial coefficient.
Now, by using the above discussed binomial theorem we have to expand the given binomial question\[{{\left( x-y \right)}^{10}}\].
\[\Rightarrow {{\left( x-y \right)}^{10}}=\sum\limits_{k=0}^{10}{\dfrac{10!}{\left( 10-k \right)!10!}}\times {{x}^{10-k}}\times {{\left( -y \right)}^{10}}\]
Now we have to expand the summation.
\[\begin{align}
  & \Rightarrow {{\left( x-y \right)}^{10}}=\dfrac{10!}{\left( 10-0 \right)!0!}.\left( {{x}^{10-0}} \right).{{\left( -y \right)}^{0}}+\dfrac{10!}{\left( 10-1 \right)!1!}.\left( {{x}^{10-1}} \right).{{\left( -y \right)}^{1}}+\dfrac{10!}{\left( 10-2 \right)!2!}.\left( {{x}^{10-2}} \right).{{\left( -y \right)}^{2}} \\
 & +\dfrac{10!}{\left( 10-3 \right)!3!}.\left( {{x}^{10-3}} \right).{{\left( -y \right)}^{3}}+\dfrac{10!}{\left( 10-4 \right)!4!}.\left( {{x}^{10-4}} \right).{{\left( -y \right)}^{4}}+....+\dfrac{10!}{\left( 10-10 \right)!10!}.\left( {{x}^{10-10}} \right).{{\left( -y \right)}^{10}} \\
\end{align}\]
Now, we have to simplify the above form.
\[\begin{align}
  & \Rightarrow {{\left( x-y \right)}^{10}}=\left( 1.{{\left( -y \right)}^{0}}.{{x}^{10}} \right)+\left( 10.{{\left( -y \right)}^{1}}.{{x}^{9}} \right)+\left( 45.{{\left( -y \right)}^{2}}.{{x}^{8}} \right)+\left( 120.{{\left( -y \right)}^{3}}.{{x}^{7}} \right) \\
 & +\left( 210.{{\left( -y \right)}^{4}}.{{x}^{6}} \right)+\left( 252.{{\left( -y \right)}^{5}}.{{x}^{5}} \right)....+\left( 1.{{\left( -y \right)}^{10}}.{{x}^{0}} \right) \\
\end{align}\]
After the simplification the above binomial expression is
\[\begin{align}
  & \Rightarrow {{\left( x-y \right)}^{10}}={{x}^{10}}-10{{x}^{10}}y+45{{x}^{8}}{{y}^{2}}-120{{x}^{7}}{{y}^{3}}+210{{x}^{6}}{{y}^{4}} \\
 & -252{{x}^{5}}{{y}^{5}}+210{{x}^{4}}{{y}^{6}}-120{{x}^{3}}{{y}^{7}}+45{{x}^{2}}{{y}^{8}}-10x{{y}^{9}}+y_{{}}^{10} \\
\end{align}\]

Note: Students should know the expansions and binomial theorem. Student should be careful with signs and calculation. Students must be careful in using the formulae of binomial theorem.
Students should not do any calculation mistakes and they must know formula like \[{{\left( a+b \right)}^{n}}=\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}.\left( {{a}^{n-k}}{{b}^{k}} \right)}\] in binomial theorem to solve questions of this kind.