1966 AHSME Problems/Problem 10
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Problem
If the sum of two numbers is 
and their product is , then the sum of their cubes is:
Solution
Let the two numbers be 
; then ![a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)[(a+b)^2 -3ab]](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/a/2/a/a2a8090d301142c423c186a1e3dc15d75d95d4b6.gif)
![= (1)[1^2 - 3 \cdot 1] = -2 \Rightarrow \mathrm{(E)}](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/1/c/7/1c7eb0a029626fe515a8afdcabac5c053f0cf5aa.gif)
.
See also
| 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 | ||
