$f\left(x\right) = x^3 + 3ax^2 +3bx + c$ has a maximum at $x=-1$ and a minimum zero at $x=1$. Find a, b and c.
$f\left(x\right) = x^3 + 3ax^2 + 3bx +c$
$f'\left(x\right) = 3x^2 + 6ax + 3b$
For maxima or minima, $f'\left(x\right) = 0$
$\implies$ $3x^2 + 6ax+3b = 0$
i.e. $x^2 + 2ax + b = 0$ $\;\; \cdots$ (1)
Now, $f\left(x\right)$ has a minimum zero at $x=1$
$\implies$ $f\left(1\right) = 0$
i.e. $1+3a+3b+c = 0$
i.e. $3a+3b+c = -1$ $\;\; \cdots$ (2)
Further, $f\left(x\right)$ has a maximum at $x=-1$ and minimum at $x=1$
$\implies$ $x = \pm 1$ are roots of quadratic equation (1)
$\therefore$ From equation (1), sum of roots $=\dfrac{2a}{1} = 1-1$
i.e. $2a = 0$ $\implies$ $a=0$
Also, from equation (1), product of roots $= \dfrac{b}{1} = 1 \times \left(-1\right)$
i.e. $b = -1$
Substituting the values of a and b in equation (2) give
$-3+c=-1$ $\implies$ $c=2$
$\therefore$ $a=0$, $b=-1$, $c=2$