Question

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

Answer 1

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

Given:
$f\left(x\right)=64x^3+\frac{1}{x^3}$
$\Rightarrow f\left(x\right)=\left(4x+\frac{1}{x}\right)\left(16x^2+\frac{1}{x^2}-4\right)$
$\Rightarrow f\left(x\right)=\left(4x+\frac{1}{x}\right)\left({\left(4x+\frac{1}{x}\right)}^2-12\right)$

$\Rightarrow f\left(\alpha \right)=\left(4\alpha +\frac{1}{\alpha }\right)\left({\left(4\alpha +\frac{1}{\alpha }\right)}^2-12\right)\mathrm{and}f\left(\beta \right)=\left(4\beta +\frac{1}{\beta }\right)\left({\left(4\beta +\frac{1}{\beta }\right)}^2-12\right)$
Since α and β are the roots of $4x+\frac{1}{x}=3$,
$4\alpha +\frac{1}{\alpha }=3\mathrm{and}4\beta +\frac{1}{\beta }=3$
$\Rightarrow f\left(\alpha \right)=3\left({\left(3\right)}^2-12\right)=-9$ and $f\left(\beta \right)=3\left({\left(3\right)}^2-12\right)=-9$
$\Rightarrow f\left(\alpha \right)=f\left(\beta \right)=-9$