1951 AMC 12 Problem 40

Question

The expression

\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^3+1)^4

1

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

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

Factor

x^3+1

and

x^3-1

Use

x^3+1=(x+1)(x^2-x+1)

and the analogous difference formula

Because

\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

1

wherever the original expression is defined. Their squared product is therefore

1.

Thus, the correct answer is C.

Problem 40 in other years