1954 AMC 12 Problem 40 — Solution | LIVE by Po-Shen Loh

LIVE

English

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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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 39#39

Problem 41#41 →

Problem 40 in Other Years

1950 AMC 12 · 1951 AMC 12 · 1952 AMC 12 · 1953 AMC 12 · 1955 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12 · 1960 AMC 12 · 1961 AMC 12 · 1962 AMC 12 · 1963 AMC 12 · 1964 AMC 12 · 1965 AMC 12 · 1966 AMC 12 · 1967 AMC 12