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Difference between revisions of "1980 AHSME Problems/Problem 27"

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== Solution ==
 
== Solution ==
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Lets set our original expression equal to <math>x</math>. So <math>\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}} = x</math>. Cubing this gives us <math>x^3 = \left(\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right)^3 = 5 + 2\sqrt{13} + 5 - 2\sqrt{13} + 3\left(\sqrt[3] {5+2\sqrt{13}}*\sqrt[3]{5-2\sqrt{13}}\right)\left(\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right) = 10 - 9x</math>
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Let <math>y = \sqrt[3] {5+2\sqrt{13}}</math>, <math>z = \sqrt[3]{5-2\sqrt{13}}</math>, and <math>x = \sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}} = y + z</math>.
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So we have <math>x^3 + 9x - 10 = 0</math>. We can easily see that 1 is a root of this polynomial. By synthetic division, the new polynomial is <math>x^2 + x + 10</math>, which has no real roots. Thus <math>\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}} = 1</math>. Since 1 is not A-D, our answer is <math>\fbox{E}</math>.
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Then <math>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</math>.
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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>.
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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.
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Therefore, <math>x = 1</math> and the answer is <math>\boxed{(\textbf{E})\ \textrm{none of these}}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 22:42, 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 (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

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