1971 AHSME Problems/Problem 22

Problem

If $w$ is one of the imaginary roots of the equation $x^3=1$ , then the product $(1-w+w^2)(1+w-w^2)$ is equal to

$\textbf{(A) }4\qquad \textbf{(B) }w\qquad \textbf{(C) }2\qquad \textbf{(D) }w^2\qquad \textbf{(E) }1$

Solution 1

Expanding the given expression yields $1+w-w^2-w-w^2+w^3+w^2+w^3-w^4=1-w^2+2w^3-w^4$ . Recalling that $w^3=1$ , we see that this expression equals $1-w^2+2-w=4-(1+w+w^2)$ . By the properties of roots of unity $\neq 1$ , $w^2+w+1=0$ , so the given expression equals $\boxed{\textbf{(A) }4}$ .

Solution 2 (not recommended)

Suppose $w=e^{i\tfrac{2\pi}3}=\cos(\tfrac{2\pi}3)+i\sin(\tfrac{2\pi}3) = \tfrac{-1+i\sqrt3}2$ . Substituting this into the given expression, we can calculate the result: \begin{align*} (1-w+w^2)(1+w-w^2) &= (1-\frac{-1+i\sqrt3}2+(\frac{-1+i\sqrt3}2)^2)(1+\frac{-1+i\sqrt3}2-(\frac{-1+i\sqrt3}2)^2) \\ &= (1-\frac{-1+i\sqrt3}{2}+\frac{1-3-2i\sqrt3}{4})(1+\frac{-1+i\sqrt3}2-\frac{1-3-2i\sqrt3}{4}) \\ &= (1-\frac{-1+i\sqrt3}{2}+\frac{-1-i\sqrt3}{2})(1+\frac{-1+i\sqrt3}2-\frac{-1-i\sqrt3}{2}) \\ &= (1-2(\frac{i\sqrt3}2))(1+2(\frac{i\sqrt3}2)) \\ &= 1^2-(i\sqrt3)^2 \\ &= 1+3 \\ &= 4 \end{align*} Thus, our answer is $\boxed{\textbf{(A) }4}$ .

1971 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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1971 AHSME Problems/Problem 22

Contents

Problem

Solution 1

Solution 2 (not recommended)

See Also