Question : If $x=\sqrt{a}+\frac{1}{\sqrt{a}}$ and $y=\sqrt{a}-\frac{1}{\sqrt{a}}, (a>0),$ then the value of $(x^{4}+y^{4}-2x^{2}y^{2})$ is:
Option 1: 16
Option 2: 20
Option 3: 10
Option 4: 5
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Correct Answer: 16
Solution :
$x=\sqrt{a}+\frac{1}{\sqrt{a}}$
Taking square on both sides,
$x^2 = a+\frac{1}{a}+2$---------------(i)
$y=\sqrt{a}-\frac{1}{\sqrt{a}}$
Taking square on both sides,
$y^2 = a+\frac{1}{a}-2$---------------(ii)
Subtracting (ii) from (i),
$x^2-y^2 = (a+\frac{1}{a}+2)-(a+\frac{1}{a}-2)$
⇒ $x^2-y^2 = 4$
Taking square on both sides,
⇒ $(x^2-y^2)^2 = 4^2$
⇒ $x^4-2x^2y^2+y^4 = 16$
Hence, the correct answer is 16.
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