Given: $x^3-a x^2+b x-c=0 ...(i)$
Let $\alpha, \beta \& \gamma$ are the roots equation (i), we get
$\begin{aligned}
& \alpha+\beta+\gamma=a \\
& \alpha \beta+\beta \gamma+\gamma \alpha=b \\
& \alpha \beta \gamma=c
\end{aligned}$
Also, given: $\alpha^3+\beta^3+\gamma^3=0$
$\begin{aligned}
& \Rightarrow(\alpha+\beta+\gamma)\left(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma-\gamma \alpha\right)+3 \alpha \beta \gamma=0 \\
& \Rightarrow a\left((\alpha+\beta+\gamma)^2-3(\alpha \beta+\beta \gamma+\gamma \alpha)\right)+3 c=0 \\
& \Rightarrow a\left[a^2-3 b\right]+3 c=0 \\
& \Rightarrow a^3-3 a b+3 c=0 \Rightarrow a^3+3 c=3 a b
\end{aligned}$