Difference between revisions of "1966 AHSME Problems/Problem 10"
(solution) |
|||
| Line 11: | Line 11: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:40, 5 July 2013
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://latex.artofproblemsolving.com/6/a/d/6ad220a107929af3055e7b259dd6e6c1f07c973f.png)
![$= (1)[1^2 - 3 \cdot 1] = -2 \Rightarrow \mathrm{(E)}$](http://latex.artofproblemsolving.com/d/9/f/d9fd586c01a1c88ca2072638dacb6a7e1b998588.png)
.
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
