If $\sin \theta +\cos \theta =a$ and ${\sin }^3\theta +{\cos }^3\theta =b,$ then we get $\lambda a^3+\mu b+vc=0$ and $\lambda ,\mu ,v$ are independent of $\theta ,$ then the value of ${\lambda }^3+{\mu }^3+v^3$ is

$\because \sin \theta +\cos \theta =a\mathrm{....}\left(i\right)$ 
On squaring both sides, we get
${\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =a^2$ 
$\sin \theta +\cos \theta =\frac{a^2-1}{2}\mathrm{........}\left(ii\right)$ 
and ${\sin }^3\theta +{\cos }^3\theta =b$  
${(\sin \theta +\cos \theta )}^3-3\sin \theta \cos \theta (\sin \theta +\cos \theta )a=b[fromeqs.(i\left)and\right(ii\left)\right]$ 
$2a^3-3a^3+3a=2b$ 
$-a^3+3a-2b=\mathrm{0......}\left(iii\right)$ 
But given that, $\lambda a^3+\mu b+vc=\mathrm{0........}\left(iv\right)$
From equations(iii) and (iv), we get
$\lambda =1,\mu =-3andv=2$ 
Now, ${\lambda }^3+{\mu }^3+v^3=1^3+{(-3)}^3+{\left(2\right)}^3=1-27+8=-18$