If $$x = p + \frac{1}{p}$$ and $$y = p - \frac{1}{p}$$ then the value of x4 - 2x2y2 + y4 = ?
A. 24
B. 4
C. 16
D. 8
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & x = p + \frac{1}{p}{\text{ }} \cr & y = p - \frac{1}{p} \cr & \therefore x + y = p + \frac{1}{p} + p - \frac{1}{p} \cr & \Leftrightarrow x + y = 2p \cr & \therefore x - y = p + \frac{1}{p} - p + \frac{1}{p} \cr & \Leftrightarrow x - y = \frac{2}{p} \cr & \therefore {x^4} - 2{x^2}{y^2} + {y^4} \cr & = {x^4} + {y^4} - 2{x^2}{y^2} \cr & = {\left( {{x^2} - {y^2}} \right)^2} \cr & = {\left[ {\left( {x + y} \right)\left( {x - y} \right)} \right]^2} \cr & = {\left( {2p \times \frac{2}{p}} \right)^2} \cr & = {\left( 4 \right)^2} \cr & = 16 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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