(c) As $f(x) = {x^3} – 3{b^2}x + 2{c^3}$ is divisible by $x – a$ and $x – b$, therefore
$f(a) = 0\,\,$

$\Rightarrow {a^3} – 3{b^2}a + 2{c^3} = 0$…..$(i)$

and $f(b) = 0$==>${b^3} – 3{b^3} + 2{c^3} = 0$…..$(ii)$

From $(ii),$ $b = c$

From $(i)$, ${a^3} – 3a{b^2} + 2{b^3} = 0$ (Putting $b = c$)

$ \Rightarrow (a – b)({a^2} + ab – 2{b^2}) = 0$

$ \Rightarrow $$a = b$ or ${a^2} + ab = 2b$

Thus $a = b = c$ or ${a^2} + ab = 2{b^2}$ and $b = c$

${a^2} + ab = 2{b^2}$is satisfied by $a = – 2b$. But $b = c$.

$\therefore$ ${a^2} + ab – 2{b^2}$ and $b = c$ is equivalent to $a = – 2b = – 2c$