Question

How do you expand the binomial ${{\left( x-y \right)}^{5}}$ ?

Answer 1

From the given question we have to expand the binomial ${{\left( x-y \right)}^{5}}$. 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 this we can expand the above binomial ${{\left( x-y \right)}^{5}}$.

Complete step by step solution:
From the given question we have to expand the binomial ${{\left( x-y \right)}^{5}}$
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 binomial theorem we have to expand the binomial ${{\left( x-y \right)}^{5}}$.
$\Rightarrow {{\left( x-y \right)}^{5}}=\sum\limits_{k=0}^{5}{\dfrac{5!}{\left( 5-k \right)!k!}.\left( {{x}^{5-k}} \right)}.{{\left( -y \right)}^{k}}$
Now we have to expand the summation.

& \Rightarrow {{\left( x-y \right)}^{5}}=\dfrac{5!}{\left( 5-0 \right)!0!}.\left( {{x}^{5-0}} \right).{{\left( -y \right)}^{0}}+\dfrac{5!}{\left( 5-1 \right)!1!}.\left( {{x}^{5-1}} \right).{{\left( -y \right)}^{1}}+\dfrac{5!}{\left( 5-2 \right)!2!}.\left( {{x}^{5-2}} \right).{{\left( -y \right)}^{2}} \\\ & +\dfrac{5!}{\left( 5-3 \right)!3!}.\left( {{x}^{5-3}} \right).{{\left( -y \right)}^{3}}+\dfrac{5!}{\left( 5-4 \right)!4!}.\left( {{x}^{5-4}} \right).{{\left( -y \right)}^{4}}+\dfrac{5!}{\left( 5-5 \right)!5!}.\left( {{x}^{5-5}} \right).{{\left( -y \right)}^{5}} \\\ \end{aligned}$$ Now, we have to simplify the above form. $$\begin{aligned} & \Rightarrow {{\left( x-y \right)}^{5}}=\left( 1.{{\left( -y \right)}^{0}}.{{x}^{5}} \right)+\left( 5.{{\left( -y \right)}^{1}}.{{x}^{4}} \right)+\left( 10.{{\left( -y \right)}^{2}}.{{x}^{3}} \right) \\\ & +\left( 10.{{\left( -y \right)}^{3}}.{{x}^{2}} \right)+\left( 5.{{\left( -y \right)}^{4}}.{{x}^{1}} \right)+\left( 1.{{\left( -y \right)}^{5}}.{{x}^{0}} \right) \\\ \end{aligned}$$ After the simplification the above binomial expression is $$\Rightarrow {{\left( x-y \right)}^{5}}={{x}^{5}}-5{{x}^{4}}y+10{{x}^{3}}{{y}^{2}}-10{{x}^{2}}{{y}^{3}}+5x{{y}^{4}}-{{y}^{5}}$$ Therefore, this is the required binomial expansion for the given binomial $${{\left( x-y \right)}^{5}}$$. **Note:** Students should know the expansions and binomial theorem. Student should be careful with signs and calculation. Students must have good knowledge in the formulae $${{\left( a+b \right)}^{n}}=\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}.\left( {{a}^{n-k}}{{b}^{k}} \right)}$$ and must not do mistakes in calculation of this formula for example in the expansion of $$ {{\left( x-y \right)}^{5}}=\sum\limits_{k=0}^{5}{\dfrac{5!}{\left( 5-k \right)!k!}.\left( {{x}^{5-k}} \right)}.{{\left( -y \right)}^{k}}$$ if we write $${{y}^{k}}$$ in the place of $${{\left( -y \right)}^{k}}$$ our whole expansion will go wrong. So, we must be careful in this aspect.