Scary-Looking Surds

Let $a=\sqrt[3]{25+22\sqrt2}$ and $b=\sqrt[3]{25-22\sqrt2}$ . Then $ab = \sqrt[3]{(25+22\sqrt2)(25-22\sqrt2)} = \sqrt[3]{25^2 - 2\cdot 22^2} = \sqrt[3]{-343} = -7$ (Honestly, I was in a car and did use my phone to calculate this and suspected $7\times 7 \times 7=343$ ). Then we have:

$\begin{aligned} (a+b)(a^2-ab+b^2) & = a^3+b^3 \\ (a+b)\left((a+b)^2-3ab\right) & = 25+22\sqrt 2 + 25-22\sqrt 2 \\ (a+b)^3+21(a+b) & = 50 & \small \color{#3D99F6} \text{Let }x = a+b \\ x^3 + 21x - 50 & = 0 & \small \color{#3D99F6} \text{By rational root theorem} \\ (x-2)\left(x^2+2x+25\right) & = 0 \\ \implies x & = \boxed 2 & \small \color{#3D99F6} \text{Note that } x^2+2x+25 = 0 \text{ has no real root.} \end{aligned}$

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Reference: Rational root theorem

Let the expression equal $x$

$\sqrt[3]{25+22\sqrt{2}} + \sqrt[3]{25-22\sqrt{2}} = x \text{ }$ (A)

And let the expressions within the 1st and 2nd cube roots equal $a$ and b respectively

$a = 25+22\sqrt{2}, b = 25-22\sqrt{2}$

Hence we can rewrite (A) as

$a^\frac{1}{3} + b^\frac{1}{3} = x$

By cubing both sides, we can simplify the expression

\(\begin{aligned}

x^3 & = a + 3a^\frac{2}{3}b^\frac{1}{3} + 3a^\frac{1}{3}b^\frac{2}{3} + b \\ & = a + b + 3a^\frac{1}{3}b^\frac{1}{3}(a^\frac{1}{3} + b^\frac{1}{3})

\end{aligned}\)

We now notice that we can replace $a^\frac{1}{3} + b^\frac{1}{3}$ with $x$ here since $x = a^\frac{1}{3} + b^\frac{1}{3}$ . Hence we have

$x^3 = a + b + 3a^\frac{1}{3}b^\frac{1}{3}x$

And since $a + b = 50$ and $ab = -343$ we have

\(\begin{aligned}

x^3 & = 50 + 3(-343)^\frac{1}{3}x \\ & = 50 + -21x

\end{aligned}\)

By solving the following cubic $x^3 + 21x - 50 = 0$ by testing small values of $x$ , we notice that $x = \boxed{2}$ .

This can be verified using Wolfram Alpha .

Note: we also have 2 imaginary solutions to the cubic above, these are extraneous solutions .