Difference between revisions of "1980 AHSME Problems/Problem 27"
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Note that <math>y^3 + z^3 = 5 + 2\sqrt{13} + 5 - 2\sqrt{13} = 10</math> and <math>yz = \sqrt[3]{\left(5 + 2\sqrt{13}\right)\left(5-2\sqrt{13}\right)} = \sqrt[3]{5^2 -\left(2\sqrt{13}\right)^2} = \sqrt[3]{-27} = -3</math>. | Note that <math>y^3 + z^3 = 5 + 2\sqrt{13} + 5 - 2\sqrt{13} = 10</math> and <math>yz = \sqrt[3]{\left(5 + 2\sqrt{13}\right)\left(5-2\sqrt{13}\right)} = \sqrt[3]{5^2 -\left(2\sqrt{13}\right)^2} = \sqrt[3]{-27} = -3</math>. | ||
− | So <math>x^3 = 10 + 3 \cdot -3 \cdot x = 10-9x</math>, that is <math>x^3- | + | So <math>x^3 = 10 + 3 \cdot -3 \cdot x = 10-9x</math>, that is <math>x^3+9x-10=0</math>. This factors to <math>(x-1)(x^2+x+10) = 0</math>, which has <math>x=1</math> as its only real solution. |
Therefore, <math>x = 1</math> and the answer is <math>\boxed{(\textbf{E})\ \textrm{none of these}}</math>. | Therefore, <math>x = 1</math> and the answer is <math>\boxed{(\textbf{E})\ \textrm{none of these}}</math>. |
Revision as of 22:45, 16 August 2025
Problem
The sum equals
Solution
Let ,
, and
.
Then .
Note that and
.
So , that is
. This factors to
, which has
as its only real solution.
Therefore, and the answer is
.
See also
1980 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.