Question

How do you factor the expression ${{x}^{3}}-{{x}^{2}}y-{{y}^{3}}+x{{y}^{2}}$?

Answer 1

Hint: In this problem we need to calculate the factors of the given equation. We can observe that the given equation is a complex equation in two variables. So, we will rearrange the terms in the given equation based on their sign i.e., we will write the terms which have positive signs are at one place and which have negative signs at one place. Now we will take appropriate terms as common from the equation and simplify it to get the required result.

Complete step by step solution:
Given equation, ${{x}^{3}}-{{x}^{2}}y-{{y}^{3}}+x{{y}^{2}}$.
Rearranging the terms in the above equation, so that all the positive terms are one place and all negative terms at one place, then the given equation is modified as
$\Rightarrow {{x}^{3}}-{{x}^{2}}y-{{y}^{3}}+x{{y}^{2}}={{x}^{3}}+x{{y}^{2}}-{{x}^{2}}y-{{y}^{3}}$
Taking $x$ as common from the first two terms and at the same time taking $-y$ as common from the last two terms, then we will get
$\Rightarrow {{x}^{3}}-{{x}^{2}}y-{{y}^{3}}+x{{y}^{2}}=x\left( {{x}^{2}}+{{y}^{2}} \right)-y\left( {{x}^{2}}+{{y}^{2}} \right)$
Again, taking ${{x}^{2}}+{{y}^{2}}$ as common from the above equation, then we will have
$\Rightarrow {{x}^{3}}-{{x}^{2}}y-{{y}^{3}}+x{{y}^{2}}=\left( {{x}^{2}}+{{y}^{2}} \right)\left( x-y \right)$
Hence the factors of the given equation ${{x}^{3}}-{{x}^{2}}y-{{y}^{3}}+x{{y}^{2}}$ are $x-y$, ${{x}^{2}}+{{y}^{2}}$.

Note: We can also follow another method to solve this problem. We will rearrange the terms based on their power, then the given equation is modified as
$\Rightarrow {{x}^{3}}-{{x}^{2}}y-{{y}^{3}}+x{{y}^{2}}={{x}^{3}}-{{y}^{3}}-{{x}^{2}}y+x{{y}^{2}}$
Using the algebraic formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ in the above equation, then we will get
$\Rightarrow {{x}^{3}}-{{x}^{2}}y-{{y}^{3}}+x{{y}^{2}}=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)-{{x}^{2}}y+x{{y}^{2}}$
Taking $xy$ as common from the last two terms in the above equation, then we will have
$\Rightarrow {{x}^{3}}-{{x}^{2}}y-{{y}^{3}}+x{{y}^{2}}=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)-xy\left( x-y \right)$
Again, taking $x-y$ as common from the above equation, then we will get
$\begin{align}
  & \Rightarrow {{x}^{3}}-{{x}^{2}}y-{{y}^{3}}+x{{y}^{2}}=\left( x-y \right)\left[ {{x}^{2}}+xy+{{y}^{2}}-xy \right] \\
 & \Rightarrow {{x}^{3}}-{{x}^{2}}y-{{y}^{3}}+x{{y}^{2}}=\left( x-y \right)\left( {{x}^{2}}+{{y}^{2}} \right) \\
\end{align}$
From both the methods we have the same result.