Which of the answers listed below is equivalent to the expression:
(xx2+1)(yy2+1)+(yx2−1)(xy2−1)
assuming, as usual, that both x and y are not zero.
- 1
- 2xy
- 2x2y2+2
- 2xy+xy2
- y2x+x2y
Solution
Here goes,
(xx2+1)(yy2+1)+(yx2−1)(xy2−1)====xyx2y2+x2+y2+1+xyx2y2−x2−y2+1xy2x2y2+2xy2x2y2+xy22xy+xy2.
The answer is (d).