If f(x)=

(a) $f(\alpha)=f(\beta)=-9$

Given:

$f(x)=64 x^{3}+\frac{1}{x^{3}}$

$\Rightarrow f(x)=\left(4 x+\frac{1}{x}\right)\left(16 x^{2}+\frac{1}{x^{2}}-4\right)$

$\Rightarrow f(x)=\left(4 x+\frac{1}{x}\right)\left(\left(4 x+\frac{1}{x}\right)^{2}-12\right)$

$\Rightarrow f(\alpha)=\left(4 \alpha+\frac{1}{\alpha}\right)\left(\left(4 \alpha+\frac{1}{\alpha}\right)^{2}-12\right)$ and $f(\beta)=\left(4 \beta+\frac{1}{\beta}\right)\left(\left(4 \beta+\frac{1}{\beta}\right)^{2}-12\right)$

Since $\alpha$ and $\beta$ are the roots of $4 x+\frac{1}{x}=3$,

$4 \alpha+\frac{1}{\alpha}=3$ and $4 \beta+\frac{1}{\beta}=3$

$\Rightarrow f(\alpha)=3\left((3)^{2}-12\right)=-9$ and $f(\beta)=3\left((3)^{2}-12\right)=-9$

$\Rightarrow f(\alpha)=f(\beta)=-9$