Question : If $x=p+\frac{1}{p}$ and $y=p-\frac{1}{p}$, then the value of $x^{4}-2x^{2}y^{2}+y^{4}$ is:
Option 1: 24
Option 2: 4
Option 3: 16
Option 4: 8
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Correct Answer: 16
Solution : $x = p+\frac{1}{p}$ $y = p-\frac{1}{p}$ Now, $x^{4}-2x^{2}y^{2}+y^{4}$ = $(x^{2}-y^{2})^{2}$ = $((x+y)(x-y))^{2}$ = $[(p+\frac{1}{p}+p-\frac{1}{p})(p+\frac{1}{p}-(p-\frac{1}{p})]^{2}$ = $[2×p×2×\frac{1}{p}]^{2}$ = 16 Hence, the correct answer is 16.
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