CTK Insights

In a related post I have shown that $\sqrt[3]{2 \pm \sqrt{5}} = \frac{1 \pm \sqrt{5}}{2}.$ Without resorting to this proof, I am going to show that $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} = 1.$

Let $a = \sqrt[3]{2 + \sqrt{5}}, b = \sqrt[3]{2 - \sqrt{5}}$ and $x = a + b.$ Then

$x^3 = (a + b)^3 = a^3 + b^3 + 3ab(a + b) = 4 -3x,$

which gives us a third degree equation in $x$: $x^3 +3x - 4 = 0.$ By direct verification $x = 1$ is one of the roots of that equation. Factoring gives $x^3 +3x - 4 = (x - 1)(x^2 + x + 4).$ The second factor which is a quadratic polynomial is always positive because $x^2 +x + 4 = (x + \frac{1}{2})^2 + 3\frac{3}{4}$ and, therefore has no real roots. However, $x$ is clearly real and is then the only real root of the cubic equation $x^3 +3x - 4 = 0,$ which is 1. Of necessity $x = 1$ and we are done.

Reference

1. C. W. Trigg, Mathematical Quickies, Dover, 1985, #35

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