If the equation ${\sin }^6\theta +{\cos }^6\theta +\lambda =0$ has real solution for $\theta$, then $\lambda$ lies in the interval

Given ${\sin }^6\theta +{\cos }^6\theta +\lambda =0$
$\begin{array}{l}\lambda =-\left({\sin }^6\theta +{\cos }^6\theta \right)a^3+b^3={\left(a+b\right)}^3-3ab\left(a+b\right)\\ =-\left[{\left({\sin }^2\theta \right)}^3+{\left({\cos }^2\theta \right)}^3\right]\\ =-\left[{\left({\sin }^2\theta +{\cos }^2\theta \right)}^3-3{\sin }^2\theta {\cos }^2\theta \left({\sin }^2\theta +{\cos }^2\theta \right)\right]\\ =-\left[1-3{\sin }^2\theta {\cos }^2\theta \right]\\ =-1+\frac{3}{4}{\left(2\sin \theta \cos \theta \right)}^2=-1+\frac{3}{4}{\sin }^{2}2\theta \end{array}$                
Take $0\le {\sin }^{2}2\theta \le 1$
$\begin{array}{l}0\le \frac{3}{4}{\sin }^{2}2\theta \le 1\\ -1\le -1+\frac{3}{4}{\sin }^{2}2\theta \le \frac{3}{4}-1\\ -1\le -1+\frac{3}{4}{\sin }^{2}2\theta \le \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\\ -1\le \lambda \le \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\Rightarrow \lambda \in \left[-1,\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right]\end{array}$