Unusual Identity

By the Triple Angle Identity for cosine, we have,

$4 {\cos}^3({\theta}) - 3\cos{\theta} = \cos{3\theta}$

$\implies 4 {\cos}^2({\theta})-3 = \dfrac{\cos{3\theta}}{\cos{\theta}}$

Therefore,

$(4 {\cos}^2({9}^{\circ})-3)\times(4 {\cos}^2({27}^{\circ})-3) = \dfrac{\cos(27^{\circ})}{\cos(9^{\circ})} \times \dfrac{\cos(81^{\circ})}{\cos{27^{\circ}}}$

$= \dfrac{\sin(90^{\circ} - 81^{\circ})}{\cos(9^{\circ})}$

$= \dfrac{\sin(9^{\circ})}{\cos(9^{\circ})} = \tan(9^{\circ})$

$\implies x^2 = \boxed{81}$