If \( x=\cos 20^{\circ} \). usi QUESTION in +10 show that \( 8 x^{3}-6 x-1=0 \)

To demonstrate that \( 8 x^{3} - 6 x - 1 = 0 \) when \( x = \cos 20^{\circ} \), we can utilize the triple angle formula for cosine: \[ \cos(3\theta) = 4\cos^3 \theta - 3\cos \theta \] By letting \( \theta = 20^{\circ} \), we have: \[ \cos(60^{\circ}) = 4\cos^3(20^{\circ}) - 3\cos(20^{\circ}) \] Since \( \cos(60^{\circ}) = \frac{1}{2} \), we can substitute this into the equation: \[ \frac{1}{2} = 4\cos^3(20^{\circ}) - 3\cos(20^{\circ}) \] Multiplying through by 2 to eliminate the fraction gives: \[ 1 = 8\cos^3(20^{\circ}) - 6\cos(20^{\circ}) \] Rearranging this results in: \[ 8\cos^3(20^{\circ}) - 6\cos(20^{\circ}) - 1 = 0 \] Thus, substituting \( x = \cos(20^{\circ}) \) leads us directly to the desired equation \( 8 x^{3} - 6 x - 1 = 0 \). So, we've shown that indeed \( 8 x^{3} - 6 x - 1 = 0 \) holds true for \( x=\cos 20^{\circ} \).