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?

I have now my own answer developed out of conversation with David Quinn, so I thank him. I publish this answer to demonstrate the calculation. It seems to coincide with the calculation of Macavity in the comments above. Wow what a monster of a question.
We have
$$x^3+ax=b$$
$$y^3+ay=c$$
Where $x^2+y^2=1$.
Adding and subtracting,
$$x^3+y^3+a(x+y)=b+c$$
$$x^3-y^3+a(x-y)=b-c$$
$$(x+y)(x^2-xy+y^2)+a(x+y)=b+c$$
$$(x-y)(x^2+xy+y^2)+a(x-y)=b-c$$
$$(x+y)(1-xy)+a(x+y)=b+c$$
$$(x-y)(1+xy)+a(x-y)=b-c$$
So
$$(x+y)(1+a-xy)=b+c$$
$$(x-y)(1+a+xy)=b-c$$
squaring,
$$(1+2xy)((1+a)^2+x^2y^2-2(a+1)xy)=(b+c)^2$$
$$(1-2xy)((1+a)^2+x^2y^2+2(a+1)xy)=(b-c)^2$$
And adding,
$$2(1+a)^2+2x^2y^2-8(a+1)x^2y^2=2(b^2+c^2)$$
So
$$(8a+6)x^2y^2=2(1+a)^2-2(b^2+c^2)$$
$$(4a+3)x^2y^2=(1+a)^2-(b^2+c^2)$$
$$x^2y^2=\frac{(1+a)^2-(b^2+c^2)}{4a+3}$$
On the other hand if we multiply the two equations together we get
$$xy(x^2y^2 +a(1+a))=bc$$
and squaring,
$$x^2y^2(x^2y^2 +a(1+a))^2=b^2c^2$$
and substituting,
$$\frac{(1+a)^2-(b^2+c^2)}{4a+3}\left(\frac{(1+a)^2-(b^2+c^2)}
{4a+3} +a(1+a)\right)^2=b^2c^2$$
which becomes,
$$[(1+a)^2-(b^2+c^2)]\left((1+a)(2a+1)^2-(b^2+c^2)\right)^2=b^2c^2(4a+3)^3$$
