1978 AMC 12 Problem 3

Attempt Problem 3 of the 1978 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1978 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

3.

For all nonzero numbers xx and yy such that x=1y,x=\frac{1}{y}, (x−1x)(y+1y) \left(x-\frac1x\right)\left(y+\frac1y\right) equals
2x22x^2
2y22y^2
x2+y2x^2+y^2
x2−y2x^2-y^2
y2−x2y^2-x^2
Answer: D
Concepts:substitutiondifference of squares
Small hint:
Use the reciprocal relation to replace 1x\frac{1}{x} and 1y\frac{1}{y}
Big hint:
The two factors become x−yx-y and x+yx+y
Solution:
Since x=1y,x=\frac{1}{y}, we also have y=1x.y=\frac{1}{x}. Thus the product is (x−y)(y+x)=x2−y2. (x-y)(y+x)=x^2-y^2.
Therefore, the correct answer is D.
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