$x =\sqrt[3]{a + \dfrac{a+1}{3}\sqrt{\dfrac{8a -1}{3}}} + \sqrt[3]{a - \dfrac{a+1}{3}\sqrt{\dfrac{8a -1}{3}}}$
$\Leftrightarrow x^3 = 2a + 3x.\sqrt[3]{a^2 - \left(\dfrac{a+1}{3}\sqrt{\dfrac{8a -1}{3}}\right)^2}$
$\Leftrightarrow x^3 = 2a + 3x.\sqrt[3]{\dfrac{1 - 6a + 12a^2 - 8a^3}{27}}$
$\Leftrightarrow x^3= 2a + 3x.\dfrac{1 - 2a}{3}$
$\Leftrightarrow x^3 = 2a + x(1 - 2a)$
$\Leftrightarrow x(x^2 -1) + 2a(x -1) = 0$
$\Leftrightarrow x(x +1)(x -1) + 2a(x -1) = 0$
$\Leftrightarrow (x -1)(x^2 + x + 2a) = 0$
$\Leftrightarrow \left[\begin{array}{l}x = 1\\x^2 + x + 2a = 0 \,\,\,\text{(vô nghiệm)}\end{array}\right.$
Vậy $x = 1$ (nguyên dương)
