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^3-9x+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.
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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 $\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$ equals

$\text{(A)} \ \frac 32 \qquad  \text{(B)} \ \frac{\sqrt[3]{65}}{4} \qquad  \text{(C)} \ \frac{1+\sqrt[6]{13}}{2} \qquad  \text{(D)}\ \sqrt[3]{2}\qquad \text{(E)}\ \text{none of these}$

Solution

Let $y = \sqrt[3] {5+2\sqrt{13}}$, $z = \sqrt[3]{5-2\sqrt{13}}$, and $x = \sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}} = y + z$.

Then $x^3 = (y+z)^3 = y^3 + 3y^2z + 3yz^2 + z^3 = y^3 + z^3 + 3yz(y+z) = y^3 + z^3 + 3yzx$.

Note that $y^3 + z^3 = 5 + 2\sqrt{13} + 5 - 2\sqrt{13} = 10$ and $yz = \sqrt[3]{\left(5 + 2\sqrt{13}\right)\left(5-2\sqrt{13}\right)} = \sqrt[3]{5^2 -\left(2\sqrt{13}\right)^2} = \sqrt[3]{-27} = -3$.

So $x^3 = 10 + 3 \cdot -3 \cdot x = 10-9x$, that is $x^3+9x-10=0$. This factors to $(x-1)(x^2+x+10) = 0$, which has $x=1$ as its only real solution.

Therefore, $x = 1$ and the answer is $\boxed{(\textbf{E})\ \textrm{none of these}}$.

See also

1980 AHSME (ProblemsAnswer KeyResources)
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

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