Not Ugly Looking After All!

Let $y$ denote the value of the expression in question, and let $x = (52 + 6\sqrt{43} )^{1/2} - (52 - 6\sqrt{43} )^{1/2}$ , and we hope to find a relationship between $x$ and $y$ .

By squaring the equation $x = (52 + 6\sqrt{43} )^{1/2} - (52 - 6\sqrt{43} )^{1/2}$ , (recall the identity, $(a-b)^2 = a^2+b^2-2ab$ ) we have

$\begin{aligned} x^2 &=& \left [ (52 + 6\sqrt{43} )^{1/2} - (52 - 6\sqrt{43} )^{1/2} \right ]^2 \\ &=& \left [ (52 + 6\sqrt{43} ) + (52 - 6\sqrt{43} ) - 2 \sqrt{(52 + 6\sqrt{43} )(52 - 6\sqrt{43} )} \right ] \\ &=& \left [ (52 + \bcancel{6\sqrt{43}} ) + (52 - \bcancel{6\sqrt{43}} ) - 2 \sqrt{52^2 - 6^2 (43)} \right ] \\ &=& \left [ 2(52) - 2 \sqrt{34^2} \right ] \\ &=& 36 \\ x &=& 6 \end{aligned}$

Note that we take the positive $x$ only because $x$ is $52 + 6\sqrt{43} > 52 - 6\sqrt{43} >0 \Rightarrow (52 + 6\sqrt{43} )^{1/2} > (52 - 6\sqrt{43} )^{1/2}\Rightarrow x> 0$ .

Similarly, raising both sides of the equation $x = (52 + 6\sqrt{43} )^{1/2} - (52 - 6\sqrt{43} )^{1/2}$ by 3, (recall the identity, $(a-b)^3 = a^3+b^3-3ab(a-b)$ ) we have

$\begin{aligned} x^3 &=& (52 + 6\sqrt{43} )^{3/2} - (52 - 6\sqrt{43} )^{3/2} - 3\sqrt{(52 + 6\sqrt{43} )(52 - 6\sqrt{43} )} \cdot x \\ 6^3 &=& y - 3 \sqrt{52^2- 6^2 \cdot 43} \cdot 6 \\ 216 &=& y - 18 \sqrt{34^2} \\ y &=& \boxed{828} \end{aligned}$

We first find the square roots, making the Ansatz $\sqrt{52+6\sqrt{43}}=a+b\sqrt{43}$ . Squaring both sides we want $a^2+43b^2=52$ and $2ab=6$ , with the positive solution $a=3,b=1$ . Thus $\sqrt{52+6\sqrt{43}}=3+\sqrt{43}$ . Likewise, $\sqrt{52-6\sqrt{43}} = -3+\sqrt{43}$ . Using the Binomial Theorem, we find $(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}$ = $(3+\sqrt{43})^3-(-3+\sqrt{43})^3$ = $3^3+3\times3^2\sqrt{43}+3\times3\times43+43\sqrt{43}-$ $( -3^3+3\times3^2\sqrt{43}-3\times3\times43+43\sqrt{43})$ $=2\times3^3+2\times3^2\times43=828$ .