Correct question is:

Prove that: 4 * sin x * sin(x + π/3) * sin(x + 2π/3) = sin 3x

Solution:

Given, 4 * sin x * sin(x + π/3) * sin(x + 2π/3)

= 4 * sin x * {sin x * cos π/3 + cos x * sin π/3} * {sin x * cos 2π/3 + cos x * sin 2π/3}

= 4 * sin x * {(sin x)/2 + (√3 * cos x)/2} * {-(sin x)/2 + (√3 *cos x)/2}

= 4 * sin x * {-(sin² x)/4 + (3 * cos² x)/4}

= sin x * {-sin² x + 3 * cos² x}

= sin x * {-sin² x + 3 * (1 - sin² x)}

= sin x * {-sin² x + 3 - 3 * sin² x}

= sin x * {3 - 4 * sin² x}

= 3* sin x - 4sin³ x

= sin 3x

So, 4 * sin x * sin(x + π/3) * sin(x + 2π/3) = sin 3x