1977 AHSME Problems/Problem 23

If the solutions of the equation $x^2+px+q=0$ are the cubes of the solutions of the equation $x^2+mx+n=0$ , then

$\textbf{(A) }p=m^3+3mn\qquad \textbf{(B) }p=m^3-3mn\qquad \textbf{(C) }p+q=m^3\qquad\\ \textbf{(D) }\left(\frac{m}{n}\right)^2=\frac{p}{q}\qquad \textbf{(E) }\text{none of these}$

Solution

Let $r_1$ and $r_2$ be the roots of the equation $x^2+mx+n=0$ . Then, $r_1^3$ and $r_2^3$ are the roots of $x^2+px+q=0$ . Applying Vieta's Formulas to the first equation, we have $r_1 + r_2 = -m$ and $r_1r_2 = n$ . Similarly, from the second equation, we have $r_1^3 + r_2^3 = -p$ and $r_1^3r_2^3 = q$ . Cubing $r_1 + r_2 = -m$ , we get $r_1^3 + r_2^3 + 3r_1r_2(r_1+r_2) = -m^3$ . Plugging in from the other equations gives $-p + 3n(-m)=-m^3$ . Rearranging we have $\boxed{\textbf{(B)} p=m^3-3mn.}$

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1977 AHSME Problems/Problem 23

Solution