Given equation is $x^3-9 x^2+23 x-15=0$ with $\alpha, \beta$ and $\gamma$ roots of equation.


Take square both sides to eq. (i),
$$
\begin{aligned}
& (\alpha+\beta+\gamma)^2=(9)^2 \\
& \alpha^2+\beta^2+\gamma^2+2(\alpha \beta+\beta \gamma+\gamma \alpha)=81 \\
& \alpha^2+\beta^2+\gamma^2+2 \alpha \beta+2 \beta \gamma+2 \gamma \alpha=81 \\
& \alpha^2+\beta^2+\gamma^2+2(23)=81 \\
& \alpha^2+\beta^2+\gamma^2+46=81 \\
& \alpha^2+\beta^2+\gamma^2=35 \quad \ldots(\text { iv }) \\
& \text { Now, } \alpha^3+\beta^3+\gamma^3=(\alpha+\beta+\gamma)\left(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma\right. \\
& -\gamma \alpha)+3 \alpha \beta \gamma
\end{aligned}
$$
$$
\begin{aligned}
& \alpha^3+\beta^3+\gamma^3=(9)(35-23)+3 \times 15 \\
& =9 \times(12)+45 \\
& =108+45=153
\end{aligned}
$$
So, option (c) is correct.