An algebra problem by Sumukh Bansal

$\begin{aligned} x & = \sqrt[3]{27+\sqrt{756}} + \sqrt[3]{27-\sqrt{756}} \\ \implies x^3 & = \left(\sqrt[3]{27+\sqrt{756}}\right)^3 + 3\left(\sqrt[3]{27+\sqrt{756}}\right)^2 \sqrt[3]{27-\sqrt{756}} + 3 \sqrt[3]{27+\sqrt{756}} \left(\sqrt[3]{27-\sqrt{756}}\right)^2 + \left(\sqrt[3]{27-\sqrt{756}}\right)^3 \\ & = 27+\sqrt{756} + 3 \sqrt[3]{27+\sqrt{756}} {\color{#3D99F6}\sqrt[3]{27^2-756}} + 3{\color{#3D99F6}\sqrt[3]{27^2-756}} \sqrt[3]{27-\sqrt{756}} + 27-\sqrt{756} \\ & = 54 + 3\left({\color{#3D99F6}\sqrt[3]{-27}} \right)\left(\sqrt[3]{27+\sqrt{756}} + \sqrt[3]{27-\sqrt{756}}\right) \\ & = 54 - 9x \end{aligned}$

$\begin{aligned} \implies x^3 + 9x - 54 & = 0 \\ (x-3)(x^2+3x+18) & = 0 \\ \implies x & = \boxed{3} \end{aligned}$

Note that $x^2+3x+18=0$ has no real root.