If $f(x)=64 x^{3}+\frac{1}{x^{3}}$ and $\alpha, \beta$ are the roots of $4 x+\frac{1}{x}=3$. Then,
(a) f(α) = f(β) = −9
(b) f(α) = f(β) = 63
(c) f(α) ≠ f(β)
(d) none of these
(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 $f(\beta)=3\left((3)^{2}-12\right)=-9$
$\Rightarrow f(\alpha)=f(\beta)=-9$
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