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$$(26+15\cdot\sqrt3)^{1/3}+(26-15\cdot\sqrt3)^{1/3}$$

I'm trying to get the result of this number. Through some algebra I found that it is close to $52^{1/3}$. Through some observation I found that it is a root of this cubic equation $x^3-3x-52=0$ and I found that the only real solution of that eq. is $4$ so now I know that the number I'm looking for is $4$ (close to my first try $52^{1/3}$). My question is, is there any algebraic process to get from the original expression and simplify it to $4$?

Thank you all very much in advance.

Eesu
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    What exactly do you want help with? How to obtain that cubic equation for $x$? Or how to solve that cubic equation for $x$? – TMM Apr 26 '13 at 22:44
  • I want to know how to get from that expression to 4. – Eesu Apr 26 '13 at 22:53

1 Answers1

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$26 + 15 \sqrt{3} = 8 + 12 \sqrt{3} + 18 + 3 \sqrt{3} = 2^3 + 3 \cdot 2^2 \sqrt{3} + 3 \cdot 2 \sqrt{3}^2 + \sqrt{3}^3 = (2 + \sqrt{3})^3.$

$26 - 15 \sqrt{3} = 8 - 12 \sqrt{3} + 1 8- 3 \sqrt{3} = 2^3 - 3 \cdot 2^2 \sqrt{3} + 3 \cdot 2 \sqrt{3}^2 - \sqrt{3}^3 = (2 - \sqrt{3})^3.$

$(26 + 15 \sqrt{3})^{\frac{1}{3}} + (26 - 15 \sqrt{3})^{\frac{1}{3}} = 2 + \sqrt{3} + 2 - \sqrt{3} = 4.$

Vladimir Reshetnikov
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    Thank you very much! It is exactly what I needed. – Eesu Apr 26 '13 at 23:03
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    +1 How did you have this idea? Did you try first something like

    \begin{equation} 26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3\left( a^{2}b+b^{3}\right) \sqrt{3} \end{equation}

    and solved the system \begin{equation} \left{ \begin{array}{c} a^{3}+9ab^{2}=26 \ a^{2}b+b^{3}=5 \end{array} \right. \end{equation}

    a solution of which is $a=2,b=1$?

     – Américo Tavares Apr 26 '13 at 23:27
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    I knew that the expected answer was 4, so I suspected that 26+15sqrt(3)=(2+nsqrt(3))^3. Then I noticed that 26-2^3=18 and 18=32sqrt(3)^2, so n=1. – Vladimir Reshetnikov Apr 26 '13 at 23:38
  • Thanks for your reply. It was a clever guess! – Américo Tavares Apr 26 '13 at 23:52