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The factors of $ {x^4} + {y^4} + {x^2}{y^2} $ are
A. $ \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

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Last updated date: 20th Jun 2024
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Answer
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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) $ .

Complete step-by-step answer:
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) $