1966 AHSME Problems/Problem 10

From AoPSWiki

Problem

If the sum of two numbers is $1$ and their product is $1$ , then the sum of their cubes is:

$\text{(A)} \ 2 \qquad \text{(B)} \ - 2 - \frac {3i\sqrt {3}}{4} \qquad \text{(C)} \ 0 \qquad \text{(D)} \ - \frac {3i\sqrt {3}}{4} \qquad \text{(E)} \ - 2$

Solution

Let the two numbers be $a,b$ ; then $a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)[(a+b)^2 -3ab]$ $= (1)[1^2 - 3 \cdot 1] = -2 \Rightarrow \mathrm{(E)}$ .

1966 AHSME (Problems)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30

Personal tools

Navigation

Search

Toolbox

1966 AHSME Problems/Problem 10

From AoPSWiki

Problem

Solution

See also