1977 AMC 12 Problem 23

Attempt Problem 23 of the 1977 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 1977 AMC 12 solutions, or check the answer key.

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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

p=m^3+3mn

p=m^3-3mn

p+q=m^3

\left(\frac mn\right)^3=\frac pq

none of these

Call the roots of the second equation

u

and

v

and apply Vieta’s formulas

Use

u^3+v^3=(u+v)^3-3uv(u+v)

Let the roots of

x^2+mx+n

u,

v.

Then

u+v=-m

and

uv=n,

while the roots of the first equation are

u^3,

v^3.

Thus

\begin{aligned} -p&=u^3+v^3\\ &=(u+v)^3-3uv(u+v)\\ &=-m^3+3mn. \end{aligned}

Therefore

p=m^3-3mn.

Therefore, the correct answer is B.

Problem 23 in other years