For all positive numbers x, y, z, the product &(x+y+z)^-1 & ×(x^-1+y^-1+z^-1) & ×(xy+yz+zx)^-1 & ×((xy)^-1+(yz)^-1+(zx)^-1) equals

1981 AMC 12 Problem 8

Attempt Problem 8 of the 1981 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 1981 AMC 12 solutions, or check the answer key.

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

For all positive numbers

x,

y,

z,

the product

\begin{aligned} &(x+y+z)^{-1}\\ &\quad\cdot(x^{-1}+y^{-1}+z^{-1})\\ &\quad\cdot(xy+yz+zx)^{-1}\\ &\quad\cdot\big((xy)^{-1}+(yz)^{-1}+(zx)^{-1}\big) \end{aligned}

equals

x^{-2}y^{-2}z^{-2}

x^{-2}+y^{-2}+z^{-2}

(x+y+z)^{-1}

\frac1{xyz}

\frac1{xy+yz+zx}

Rewrite each sum of reciprocals over a common denominator

The two reciprocal sums introduce factors

xy+yz+zx

and

x+y+z

We have

x^{-1}+y^{-1}+z^{-1}=\frac{xy+yz+zx}{xyz}

and

\begin{aligned} &(xy)^{-1}+(yz)^{-1}\\ &\quad+(zx)^{-1} =\frac{x+y+z}{xyz}. \end{aligned}

All symmetric sum factors cancel, leaving

\frac{1}{x^2y^2z^2}=x^{-2}y^{-2}z^{-2}.

Therefore, the correct answer is A.