Simplifying roots 1

We note that $\begin{aligned} \left( 1 + \frac{1}{\sqrt{3}} \right)^3 = 1 + \frac{3}{\sqrt{3}} + \frac{3}{3} + \frac{1}{3\sqrt{3}} = 2 + \frac{10}{3\sqrt{3}} \end{aligned}$ . Similarly, $\begin{aligned} \left( 1 - \frac{1}{\sqrt{3}} \right)^3 = 2 - \frac{10}{3\sqrt{3}} \end{aligned}$ .

Therefore,

$\begin{aligned} x & = \sqrt [3] {2+\frac{10}{3\sqrt{3}}} + \sqrt [3] {2-\frac{10}{3\sqrt{3}}} \\ & = \sqrt [3] {\left( 1 + \frac{1}{\sqrt{3}} \right)^3} + \sqrt [3] {\left( 1 - \frac{1}{\sqrt{3}} \right)^3} \\ & = 1 + \frac{1}{\sqrt{3}} + 1 - \frac{1}{\sqrt{3}} \\ & = \boxed{2} \end{aligned}$

Let $x = \sqrt[3]{2 + \frac{10}{3\sqrt{3}}} + \sqrt[3]{2 - \frac{10}{3\sqrt{3}}}$ . $\begin{aligned} x^{3} & = 2 + \frac{10}{3\sqrt{3}} + 2 - \frac{10}{3\sqrt{3}} + 3\sqrt[3]{2 + \frac{10}{3\sqrt{3}}}\sqrt[3]{2 - \frac{10}{3\sqrt{3}}}(\sqrt[3]{2 + \frac{10}{3\sqrt{3}}} + \sqrt[3]{2 - \frac{10}{3\sqrt{3}}})\\ & = 4 + 3\sqrt[3]{2^{2} -( \frac{10}{3\sqrt{3}})^{2}}x \\ & = 4 + 3\sqrt[3]{4 - \frac{100}{27}}x \\ & = 4 + 3\sqrt[3]{\frac{8}{27}}x \\ & = 4 + 2x \end{aligned}$ .

Now, $\begin{aligned} x^{3} - 2x - 4 & = 0 \\ (x - 2)(x^{2} + 2x + 2) &= 0 \end{aligned}$ .

Either, $x = 2$ or $x^{2} + 2x + 2 = 0$ . Now, $\begin{aligned} x^2 + 2x + 2 & = 0 \\ (x + 1)^{2} & = -1 \end{aligned}$ has no real solutions for $x$ . Also, for a quadratic equation $ax^{2} + bx + c = 0$ , where $a = 1,\ b = c = 2, \ \Delta\ =\ b^{2} - 4ac = 2^{2} - 4*1*2 = -4 < 0$ . Hence, $x = \boxed{2}$

just take the equation equal to x and then take cube both the side.. it will become a cubic in x whose one solution is clearly visible.