Let , where p is a constant, then  at x = 0 is

Formula used:

Calculation:

f(x) = x³(- 1 × p³ -  0 × p²) - sinx(6 × p³ - 0 × p) + cosx(6p² + p)

⇒ f(x) =  - p3x3 - sinx.6p3 + cosx(6p2 + p)

⇒ f'(x) = -3p³x² - cosx.6p³ - sinx(6p² + p)

⇒ f''(x) = -6p3x + sinx.6p3 - cosx.(6p2 + p)

⇒ f'''(x) = -6p3 + cosx.6p3 + sinx.(6p2 + p)

Put x = 0 in above equation, we will get

 f'''(0) = -6p3 + cos0.6p3 + sin0.(6p2 + p)

⇒ f'''(0) = -6p3 + 6p3

⇒ f'''(0) = 0

We can see that  at x = 0 is independent of p.