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# The factors of ${x^4} + {y^4} + {x^2}{y^2}$ areA. $\left( {{x^2} + {y^2}} \right)\left( {{x^2} + {y^2} - xy} \right)$ B. $\left( {{x^2} + {y^2}} \right)\left( {{x^2} - {y^2}} \right)$ C. $\left( {{x^2} + {y^2} + xy} \right)\left( {{x^2} + {y^2} - xy} \right)$ D.Factorization is not possible

Last updated date: 20th Jun 2024
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Hint: The factors can be obtained by the method of completing the squares. The formula to be used are ${a^2} + {b^2} + 2ab = {\left( {a + b} \right)^2}$ and ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ .

The given expression is
$E = {x^4} + {y^4} + {x^2}{y^2} \cdots \left( 1 \right)$
In order to factorise the expression we will use, the method of completing the squares. After comparing the terms with ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ in the formula, the 3 terms are,
${a^2} = {x^4}$
Therefore, $a = {x^2}$
${b^2} = {y^4}$
Therefore, $b = {y^2}$
But the 3-rd term is ${x^2}{y^2}$ . It should have been $2{x^2}{y^2}$ but the case is not like that.
In order to make it a perfect square, add and subtract ${x^2}{y^2}$ in equation (1). The equation becomes as,
$E = {x^4} + {y^4} + {x^2}{y^2} + \left( {{x^2}{y^2}} \right) - \left( {{x^2}{y^2}} \right) \cdots \left( 2 \right)$
The two ${x^2}{y^2}$ will combine to give $2{x^2}{y^2}$ which is our requirement in completing the square.
The equation (2) becomes as,
$E = {x^4} + {y^4} + 2{x^2}{y^2} - {x^2}{y^2} \cdots \left( 3 \right)$
The 3 terms ${x^4}$ , ${y^4}$ and $2{x^2}{y^2}$ form a perfect square and satisfies the identity ${a^2} + {b^2} + 2ab = {\left( {a + b} \right)^2}$ . Now, equation (3) becomes as,
${\left( {{x^2} + {y^2}} \right)^2} - {x^2}{y^2} \cdots \left( 4 \right)$
Equation (4) can be written as
${\left( {{x^2} + {y^2}} \right)^2} - {\left( {xy} \right)^2} \cdots \left( 5 \right)$
We can use another identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ to factorize equation (5). Here $a = {x^2} + {y^2}$ and . Now equation (5) can be written as,
$E = \left( {{x^2} + {y^2} + xy} \right)\left( {{x^2} + {y^2} - xy} \right)$
Therefore, the factors of ${x^4} + {y^4} + {x^2}{y^2}$ are $\left( {{x^2} + {y^2} + xy} \right)$and $\left( {{x^2} + {y^2} - xy} \right)$ .

So, the correct answer is “Option C”.

Note: The important steps are
The use of the method of completing the squares. Just by adding or subtracting the suitable terms a perfect square can be made as ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ .
For instance, the expression $e = {a^2} + {b^2} + ab$ can be made a perfect square by adding and subtracting as,
$e = {a^2} + {b^2} + ab + ab - ab$
$e = {a^2} + {b^2} + 2ab - ab \\ e = {\left( {a + b} \right)^2} - ab \\$
The use of ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ . For instance, the factors of the expression $f = 81{x^2} - 36{y^2}$ .
Here, $a = 9x$ and $b = 6y$ .
The expression becomes as
$f = \left( {9x + 6y} \right)\left( {9x - 6y} \right)$