Question

Factorize the given expression and choose the right option: ${x^4} + {y^4} - 27{x^2}{y^2}$
$
  (a){\text{ }}\left( {{x^2} - 5xy + {y^2}} \right)\left( {{x^2} + 5xy - {y^2}} \right) \\
  (b){\text{ }}\left( {{x^2} + xy - {y^2}} \right)\left( {{x^2} - 5y - {y^2}} \right) \\
  (c){\text{ }}\left( {{x^2} + xy - {y^2}} \right)\left( {{x^2} - 5x - {y^2}} \right) \\
  (d){\text{ }}\left( {{x^2} + 5xy - {y^2}} \right)\left( {{x^2} - 5xy - {y^2}} \right) \\
 $

Answer 1

Hint – In order to factorize the given expression add and subtract $5{x^3}y$ and $5{x}{y^3} $ to the given expression and break $(-27{x^2} {y^2})$ into three terms i.e. $(-{x^2} {y^2}, -25 {x^2} {y^2} \text{and} -{x^2}{y^2}) $. This will help get the right option.

Complete step-by-step solution -
Given equation is
${x^4} + {y^4} - 27{x^2}{y^2}$
Now we have to factorize this.
So add and subtract by $5x{y^3}$ and $5{x^3}y$.
And break $(-27{x^2} {y^2})$ into three terms i.e. $(-{x^2} {y^2}, -25 {x^2} {y^2} \text{and} -{x^2}{y^2}) $.
So the above equation is written as
$ \Rightarrow {x^4} - 5{x^3}y - {x^2}{y^2} + 5{x^3}y - 25{x^2}{y^2} - 5x{y^3} - {x^2}{y^2} + 5x{y^3} + {y^4}$
Now take common ${x^2}$ from first three terms, $5xy$ from next three terms and ${-y^2}$ from last three terms we have,
$ \Rightarrow {x^2}\left( {{x^2} - 5xy - {y^2}} \right) + 5xy\left( {{x^2} - 5xy - {y^2}} \right) - {y^2}\left( {{x^2} - 5xy - {y^2}} \right)$
Now take $\left( {{x^2} - 5xy - {y^2}} \right)$ common we have,
$ \Rightarrow \left( {{x^2} + 5xy - {y^2}} \right)\left( {{x^2} - 5xy - {y^2}} \right)$
So this is the required factorization of the given equation.
Hence option (D) is correct.

Note – Factorization consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind. The addition and subtraction of $5{x^3}y$ was simply to make terms common so that simplification could be implemented.