Question

Find the value of following expression:
\[\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)\]
(a) \[{{x}^{3}}+{{y}^{3}}\]
(b) \[{{x}^{2}}+{{y}^{2}}\]
(c) \[{{x}^{3}}-{{y}^{3}}\]
(d) \[{{x}^{2}}-{{y}^{2}}\]

Answer 1

Hint: We solve this problem by using the multiplication by term to term.
We multiply each term of the first bracket with each term of the second bracket so that we can cancel out the common terms if any and the terms of the same variable to get the required result.
In this process, we need to keep in mind that we can add or subtract the terms of the same variable.

Complete step by step answer:
We are asked to find the value of \[\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)\]
Let us assume that the given product as
\[\Rightarrow A=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)\]
Now, let us multiply the each term in the first bracket with the each term in the second bracket then we get
\[\begin{align}
  & \Rightarrow A=x\left( {{x}^{2}}+xy+{{y}^{2}} \right)-y\left( {{x}^{2}}+xy+{{y}^{2}} \right) \\
 & \Rightarrow A={{x}^{3}}+{{x}^{2}}y+x{{y}^{2}}-y{{x}^{2}}-x{{y}^{2}}-{{y}^{3}} \\
\end{align}\]
Now, let us cancel out the terms in the above equation then we get
\[\Rightarrow A={{x}^{3}}-{{y}^{3}}\]

Therefore we can conclude that the value of given product as
\[\therefore \left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)={{x}^{3}}-{{y}^{3}}\]
So, option (c) is the correct answer.

Note:
We can solve this problem in another method also.
We have the standard formula of difference of cubes of two numbers as
\[\Rightarrow {{x}^{3}}-{{y}^{3}}={{\left( x-y \right)}^{3}}+3xy\left( x-y \right)\]
Now, by taking the common term out in the RHS of above equation we get
\[\Rightarrow {{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left[ {{\left( x-y \right)}^{2}}+3xy \right]\]
We know that the formula of square of difference of two numbers as
\[{{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}\]
By using this formula in above equation we get
\[\begin{align}
  & \Rightarrow {{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}-2xy+{{y}^{2}}+3xy \right) \\
 & \Rightarrow {{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right) \\
\end{align}\]
Therefore we can conclude that the value of given product as
\[\therefore \left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)={{x}^{3}}-{{y}^{3}}\]
So, option (c) is the correct answer.
Here, we can also confirm directly the value of given product by using the standard formula of difference of cubes of two numbers that is
\[\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)={{x}^{3}}-{{y}^{3}}\]
This is the direct formula we have in algebra
Therefore we can conclude that the value of given product as
\[\therefore \left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)={{x}^{3}}-{{y}^{3}}\]
So, option (c) is the correct answer.