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1951 AMC 12 Problem 40

Attempt Problem 40 of the 1951 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 1951 AMC 12 solutions, or check the answer key.

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40.

The expression ((x+1)2(x2−x+1)2(x3+1)2)2⋅((x−1)2(x2+x+1)2(x3−1)2)2 \begin{aligned} &\left(\frac{(x+1)^2(x^2-x+1)^2}{(x^3+1)^2}\right)^2\\ &\quad{}\cdot \left(\frac{(x-1)^2(x^2+x+1)^2}{(x^3-1)^2}\right)^2 \end{aligned} equals:

(x+1)4(x+1)^4

(x3+1)4(x^3+1)^4

11

[(x3+1)(x3−1)]2\big[(x^3+1)(x^3-1)\big]^2

[(x3−1)2]2\big[(x^3-1)^2\big]^2

Answer: C
Concepts:sum and difference of cubesalgebraic manipulation
Difficulty rating: 1400
Small Hint:

Factor x3+1x^3+1 and x3−1x^3-1

Big Hint:

Use x3+1=(x+1)(x2−x+1)x^3+1=(x+1)(x^2-x+1) and the analogous difference formula

Solution:

Because x3+1=(x+1)(x2−x+1),x3−1=(x−1)(x2+x+1), \begin{aligned} x^3+1&=(x+1)(x^2-x+1),\\ x^3-1&=(x-1)(x^2+x+1), \end{aligned} each fraction inside parentheses equals 11 wherever the original expression is defined. Their squared product is therefore 1.1.

Thus, the correct answer is C.

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