2150.72 – Find the Equivalent

Which of the answers listed below is equivalent to the expression:

(x2+1x)(y2+1y)+(x2−1y)(y2−1x)\left( \dfrac{x^2 + 1}{x} \right) \left( \dfrac{y^2 + 1}{y} \right) + \left( \dfrac{x^2 - 1}{y} \right) \left( \dfrac{y^2 - 1}{x} \right)

assuming, as usual, that both xx and yy are not zero.

  1. 11

  2. 2xy2 x y

  3. 2x2y2+22 x^2 y^2 + 2

  4. 2xy+2xy2 x y + \dfrac{2}{x y}

  5. 2xy+2yx\dfrac{2x}{y} + \dfrac{2y}{x}

Solution

Here goes,

(x2+1x)(y2+1y)+(x2−1y)(y2−1x)=x2y2+x2+y2+1xy+x2y2−x2−y2+1xy=2x2y2+2xy=2x2y2xy+2xy=2xy+2xy\begin{aligned}\left( \dfrac{x^2 + 1}{x} \right) \left( \dfrac{y^2 + 1}{y} \right) + \left( \dfrac{x^2 - 1}{y} \right) \left( \dfrac{y^2 - 1}{x} \right) &= \dfrac{x^2 y^2 + x^2 + y^2 + 1}{xy} + \dfrac{x^2 y^2 - x^2 - y^2 + 1}{xy} \\[1em]&= \dfrac{2 x^2 y^2 + 2}{xy} \\[1em]&= \dfrac{2 x^2 y^2}{xy} + \dfrac{2}{xy} \\[1em]&= 2 x y + \dfrac{2}{xy}\end{aligned}

The answer is d.