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: (1w+w2)(1+ww2)=(11+i32+(1+i32)2)(1+1+i32(1+i32)2)=(11+i32+132i34)(1+1+i32132i34)=(11+i32+1i32)(1+1+i321i32)=(12(i32))(1+2(i32))=12(i3)2=1+3=4 Thus, our answer is $\boxed{\textbf{(A) }4}$.

See Also

1971 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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