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Prove that $\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} = 3$ without using Cardano's formula. (Hint, what is $(3\pm \sqrt{13})^3$

I have that $$(3 + \sqrt{13})^3 = 144 + 40 \sqrt{13} $$ and $$(3 - \sqrt{13})^3 = 144 - 40 \sqrt{13} $$

A cursory look into Bombelli's method led me to the following system of equations:

$$\sqrt[3]{18 + \sqrt{325}} = a + b^{1/2}$$ $$\sqrt[3]{18 - \sqrt{325}} = a - b^{1/2}$$

I am unsure how to solve this system of equations without making a mess of the radicals...I know however that the given cube roots on the LHS of the above system are solutions to the cubic $x^3 + 3x = 36 $

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Since $(3\pm\sqrt{13})^3=144\pm40\sqrt{13}=8(18\pm5\sqrt{13}),$

$\sqrt[3]{18\pm\sqrt{325}}=\sqrt[3]{18\pm5\sqrt{13}}=\dfrac{3\pm\sqrt{13}}2$.

Therefore, $\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}}=\dfrac{3+\sqrt{13}}2+\dfrac{3-\sqrt{13}}2=3.$

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  • $\begingroup$ Could you explain how in the second line the expression simplifies to (3 +- 13^(1/2))/2 ? $\endgroup$ – kt046172 Jan 16 at 20:07
  • $\begingroup$ Take the cube root of both sides of $(3\pm\sqrt{13})^3=8(18\pm5\sqrt{13})$ and then divide both sides by $2$ $\endgroup$ – J. W. Tanner Jan 16 at 20:14
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Let $$\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}}=x$$ Thus, $$18 + \sqrt{325}+ 18 - \sqrt{325}+3\left(\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}}\right)\sqrt[3]{18 + \sqrt{325}}\sqrt[3]{18 - \sqrt{325}}=x^3$$ or $$36+3x\cdot(-1)=x^3$$ or $$x^3-3x^2+3x^2-9x+12x-36=0$$ or $$(x-3)(x^2+3x+12)=0.$$ Can you end it now?

Also, by your hint: $$\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} =\frac{3}{2}+\frac{1}{2}\sqrt{13}+\frac{3}{2}-\frac{1}{2}\sqrt{13}=3.$$

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Hint, a link to a nice resource and to complement Michael's answer. There is this famous statement

If $a+b+c=0$ then $a^3+b^3+c^3=3abc$

source, Mathematical Olympiad Treasures, by Titu Andreescu and Bogdan Enescu, first chapter is available in preview, the proof is there.


As a result $$x=\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} \Rightarrow \\ x-\sqrt[3]{18 + \sqrt{325}} - \sqrt[3]{18 - \sqrt{325}} =0 \Rightarrow \\ x^3- (18 + \sqrt{325}) - (18 - \sqrt{325}) = 3x\sqrt[3]{(18 + \sqrt{325})(18 - \sqrt{325})} \Rightarrow \\ x^3-36=-3x$$ which has $3$ as the only real solution.

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