1954 AMC 12 Problem 40

Attempt Problem 40 of the 1954 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 1954 AMC 12 solutions, or check the answer key.

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40.

If

\left(a+\dfrac1a\right)^2=3,

then

a^3+\dfrac1{a^3}

equals:

\dfrac{10\sqrt3}{3}

3\sqrt3

0

7\sqrt7

6\sqrt3

Answer: C

Concepts:algebraic manipulation sum and difference of cubes substitution

Difficulty rating: 1670

Small hint:

Set

t=a+\frac{1}{a}

, so the given condition is

t^2=3

Big hint:

Use

a^3+a^{-3}=t^3-3t

Solution:

Let

t=a+\frac{1}{a}.

The standard cubic identity gives

a^3+\frac1{a^3}=t^3-3t=t(t^2-3).

Since

t^2=3,

this expression is

0.

Thus, the correct answer is C.

Problem 40 in other years