If the function f: R → A defined as f (x) = ^ -1 ( 2x^ 3 1 + x^ 6 )...

Let, $x^{3} = t \in \mathbb{R}$
Let, $y = \dfrac{2t}{1 + t^{2}}$

$⇒ yt^{2} - 2t + y = 0$
$∵ t \in \mathbb{R}$, hence $D \ge 0$
$⇒ \left(-2\right)^{2} - 4y^{2} ≥ 0 ⇒ y \in \left[-1,1\right]$
$\tan^{-1} y \in \left[\tan^{-1} \left(-1\right), \tan^{-1} \left(1\right)\right]$

Hence, the range of $f\left(x\right)\in \left[- \dfrac{\pi}{4}, \dfrac{\pi}{4}\right]≡ A$