Eliminating $\theta$ from $\cos^3\theta +a\cos\theta =b$ and $\sin^3\theta +a\sin\theta =c$ 
Eliminate $\theta$ from the equations.
$$\cos^3\theta +a\cos\theta =b$$
$$\sin^3\theta +a\sin\theta =c$$

Can anyone solve this question?

$\def\¿{\mathcal}$
Maybe I am complicating this, yet another method.
I will use these identities:
$\cos^6x+\sin^6x=1-\frac34\sin^22x$
$\cos^6x-\sin^6x=\cos 2x(1-\frac14\sin^22x)$
$\cos^4x+\sin^4x=1-\frac12\sin^22x$
$\cos^4x-\sin^4x=\cos2x$
Let 
$c=\cos x$, 
$s=\sin x$,
$\¿ C=\cos 2x$,
$\¿S=\sin2x$
I use $d$ instead of $c$ in the original problem to avoid confusion.
Now we have $$c^3+ac=b\tag{1}$$ $$s^3+as=d\tag{2}$$

$\small\mathit{(1)^2+(2)^2}$, $$c^6+s^6+2a(c^4+s^4)+a^2(c^2+s^2)=b^2+d^2$$ $$\left(1-\frac34\¿S^2\right)+2a\left(1-\frac12\¿S^2\right)+a^2=b^2+d^2$$ $$\¿S^2=\frac{4\left[\left(a+1\right)^2-(b^2+d^2)\right]}{(3+4a)}$$

$\small\mathit{(1)^2-(2)^2}$, $$c^6-s^6+2a(c^4-s^4)+a^2(c^2-s^2)=b^2-d^2$$ $$\¿C\left(1-\frac14\¿S^2\right)+2a\¿C+a^2\¿C=b^2-d^2$$ $$\¿C\left[(a+1)^2-\frac14\¿S^2\right]=b^2-d^2$$ $$\¿C=\frac{(3+4a)(b^2-d^2)}{2(a+1)^2(2a+1)+(b^2+d^2)}$$ Now we can use $$\¿C^2+\¿S^2=1$$
