Problem: If xy=bx y=bxy=b and 1x2+1y2=a\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}=ax21​+y21​=a, then (x+y)2(x+y)^{2}(x+y)2 equals:
Answer Choices:
A. (a+2b)2(a+2 b)^{2}(a+2b)2
B. a2+b2a^{2}+b^{2}a2+b2
C. b(ab+2)b(a b+2)b(ab+2)
D. ab(b+2)\mathrm{ab}(\mathrm{b}+2)ab(b+2)
E. 1a+2b\dfrac{1}{a}+2 ba1​+2b Solution:
1x2+1y2=y2+x2x2y2−a\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}=\dfrac{y^{2}+x^{2}}{x^{2} y^{2}} - a x21​+y21​=x2y2y2+x2​−a
∴a=x2+y2b2 or x2+y2=ab2\therefore a=\dfrac{x^{2}+y^{2}}{b^{2}} \text{ or } x^{2}+y^{2}=a b^{2} ∴a=b2x2+y2​ or x2+y2=ab2
(x+y)2=x2+2xy+y2(x+y)^{2}=x^{2}+2 x y+y^{2} (x+y)2=x2+2xy+y2
=ab2+2b=b(ab+2)=a b^{2}+2 b=b(a b+2) =ab2+2b=b(ab+2)