| If \[x=p+\frac{1}{p}\]and \[y=p-\frac{1}{p},\]then value of \[{{x}^{4}}-2{{x}^{2}}{{y}^{2}}+{{y}^{4}}\] is |
A) 24
B) 4
C) 16
D) 8
Correct Answer: C
Solution :
| Given, \[x=p+\frac{1}{p}\]and \[y=p-\frac{1}{p}\] |
| \[{{x}^{2}}-2{{x}^{2}}{{y}^{2}}+{{y}^{4}}={{[({{x}^{2}}-{{y}^{2}})]}^{2}}\] |
| \[={{[(x+y)(x-y)]}^{2}}\] |
| \[={{\left[ \left( p+\frac{1}{p}+p-\frac{1}{p} \right)\left( p+\frac{1}{p}-p+\frac{1}{p} \right) \right]}^{2}}\] |
| [putting the value of a and y] |
| \[={{\left[ (2p)\left( \frac{2}{p} \right) \right]}^{2}}=4{{p}^{2}}\times \frac{4}{{{p}^{2}}}=4\times 4=16\] |
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