The sum of the solutions x ∈ R of the equation 3 cos 2 x + cos 3 2 x cos 6 x - sin 6 x = x 3 - x 2 + 6 is

Given: $\frac{3\cos 2x+{\cos }^{3}2x}{{\cos }^{6}x-{\sin }^{6}x}=x^3-x^2+6$

$\Rightarrow \frac{\cos 2x\left(3+{\cos }^{2}2x\right)}{{\left({\cos }^{2}x\right)}^3-{\left({\sin }^{2}x\right)}^3}=x^3-x^2+6$

$\Rightarrow \frac{\cos 2x\left(3+{\cos }^{2}2x\right)}{\left({\cos }^{2}x-{\sin }^{2}x\right)\left({\cos }^{4}x+{\sin }^{4}x+{\sin }^{2}x{\cos }^{2}x\right)}=x^3-x^2+6$

$\Rightarrow \frac{\left(3+{\cos }^{2}2x\right)}{{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x+{\sin }^{2}x{\cos }^{2}x}=x^3-x^2+6$

$\Rightarrow \frac{\left(3+{\cos }^{2}2x\right)}{1-{\sin }^{2}x{\cos }^{2}x}=x^3-x^2+6$

$\Rightarrow \frac{\left(3+{\cos }^{2}2x\right)}{1-{\displaystyle \frac{{\left(2\sin x\cos x\right)}^2}{4}}}=x^3-x^2+6$

$\Rightarrow 4\left(\frac{3+{\cos }^{2}2x}{4-{\displaystyle {\sin }^{2}2x}}\right)=x^3-x^2+6$

$\Rightarrow 4\left(\frac{3+1-{\sin }^{2}2x}{4-{\displaystyle {\sin }^{2}2x}}\right)=x^3-x^2+6$

$\Rightarrow 4=x^3-x^2+6$

$\Rightarrow x^3-x^2+2=0$

$\Rightarrow \left(x+1\right)\left(x^2-2x+2\right)=0$

$\Rightarrow \left(x+1\right)\left({\left(x-1\right)}^2+1\right)=0$

So, the sum of real roots is $-1$.