Given:

f(x) = x³+6x²+15x-12.

f’(x) = 3x²+12x+15

f’(x) = 3x²+12x+12+3

f’(x) = 3(x²+4x+4)+3

f’(x) = 3(x+2)²+3

As square is a positive number

∴ f’(x) will be always positive for every real number

Hence f’(x) >0 for all x ϵ R

∴ f(x) is strictly increasing.