I can factor it?

$\cos^2x = \dfrac{1+\cos2x}{2}$

$\cos^22x = \dfrac{1+\cos4x}{2}$

$\cos^23x = \dfrac{1+\cos6x}{2}$

So we can modify the L.H.S as

$\cos^2x + \cos^22x + \cos^23x - 1= \dfrac{1+\cos2x}{2} + \dfrac{1+\cos4x}{2} + \dfrac{1+\cos6x}{2} - 1 = \dfrac{\cos2x + \cos4x + \cos6x + 1}{2}$

$\cos6x + 1 = 2\cos^23x$

$\cos2x + \cos4x = 2 \cos x \cos3x$

$L.H.S = \dfrac{(2\cos3x)(\cos3x + \cos x)}{2} = \dfrac{(\cos3x)(2\cos x \cos2x)}{1}$

$L.H.S = 2\displaystyle \prod_{k=1}^{3}\cos kx$

$a = 2 \text{ \& } b = 3$

$2a + b = \huge\color{#D61F06}{\boxed{7}}$

$\begin{aligned} \cos^2 (x) + \cos^2 (2x) + \cos^2 (3x) -1 & = \cos^2 (x) + [2\cos^2 (x) -1]^2 + [4\cos^2 (x) - 3\cos^2 (x)]^2 -1 \\ & = y^2 + 4y^4 -4y^2 + 1 + 16y^6-24y^4 + 9y^2 - 1 \quad \quad \small \color{#3D99F6}{\text{Let }y = \cos x} \\ & = 16y^6 - 20y^3 + 6y \\ & = 2y(8y^5-10y^2+3y) \\ & = 2y(2y^2-1)(4y^3-3y) \\ & = 2\cos(x) \cos(2x) \cos(3) \\ & = 2\prod_{k=1}^3 \cos{kx} \end{aligned}$

$\Rightarrow 2a+b = 2(2) + 3 = \boxed{7}$ .