Question

Factorize the given expression \[\left( {{x}^{2}}-2xy+{{y}^{2}} \right)-{{z}^{2}}\]

Answer 1

Hint: In the given expression, first take the expression given inside the bracket and apply the basic formula. Now simplify the expression again by using the identity \[\left( {{a}^{2}}-{{b}^{2}} \right)\].

Complete step-by-step answer:
Let us take the given expression,
 \[\left( {{x}^{2}}-2xy+{{y}^{2}} \right)-{{z}^{2}}.........(1)\]
We know that \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\].
Similarly, \[\left( {{x}^{2}}-2xy+{{y}^{2}} \right)={{\left( x-y \right)}^{2}}\].
Thus the expression \[\left( {{x}^{2}}-2xy+{{y}^{2}} \right)-{{z}^{2}}\] becomes \[{{\left( x-y \right)}^{2}}-{{z}^{2}}\].
Now look at the expression, \[{{\left( x-y \right)}^{2}}-{{z}^{2}}......(2)\]
This is of the form \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\].
Thus let us convert the equation (2).
\[{{\left( x-y \right)}^{2}}-{{z}^{2}}=\left[ \left( x-y \right)+z \right]\left[ \left( x-y \right)-z \right]\]
Open the brackets and simplify.
\[\left[ x-y+z \right]\left[ x-y-z \right]\]
Thus we have factorized, \[\left( {{x}^{2}}-2xy+{{y}^{2}} \right)-{{z}^{2}}=\left[ x-y+z \right]\left[ x-y-z \right]\].
This cannot be factored further.
\[\therefore \left( {{x}^{2}}-2xy+{{y}^{2}} \right)-{{z}^{2}}=\left( x-y+z \right)\left( x-y-z \right)\].

Note: You can also solve it by taking common factors.
\[\begin{align}
  & \left( {{x}^{2}}-2xy+{{y}^{2}} \right)-{{z}^{2}} \\
 & \left[ x\left( x-y \right)-y\left( x-y \right) \right]-{{z}^{2}} \\
 & \Rightarrow \left[ \left( x-y \right)\left( x-y \right) \right]-{{z}^{2}} \\
 & ={{\left( x-y \right)}^{2}}-{{z}^{2}} \\
\end{align}\]
Then apply \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\].
\[={{\left( x-y \right)}^{2}}-{{z}^{2}}=\left( x-y-z \right)\left( x+y-z \right)\].