Question

Let $f (x) =x^{3} +x+1$, suppose $p(x)$ be another cubic polynomial such that $p(0) =-1$ & zero’s of $p(x)$ are the square of the roots of $f (x) =0 $, then value of $p (9)$ equals

• A. $1331$
• B. $1024$
• C. $961$
• D. $899$

Answer 1

Answer: (D)

$f (x) = x^{3}+x+1 = (x- \alpha) (x -\beta) (x- \gamma)\, \dots(A)$
where $\alpha, \beta, \gamma$ are roots of $f (x) =0$
$\therefore \alpha \beta \gamma =-1$
Now, roots of $p (x) =0$, are square of the roots of
$f (x) =0$ i.e., roots of $p (x) =0$ are $\alpha^{2}, \beta^{2}, \gamma^{2}$
$\therefore p (x) = \lambda (x-\alpha^{2}) (x-\beta^{2}) (x-\gamma^{2}) \, \dots (*)$
$\therefore p (0) = \lambda (-\alpha^{2} \beta^{2} \gamma^{2})$
$\Rightarrow 1=\lambda$ (as $\alpha \beta \gamma=-1$ & $p (0)=-1)$
Now $p(x) = (x-\alpha^{2})(x-\beta^{2}) (x-\gamma^{2})$ (using (*))
$\therefore p (x^{2}) = (x^{2} -\alpha^{2}) (x^{2}-\beta^{2}) (x^{2}-\gamma^{2})$
$=(x-\alpha) (x-\beta)(x-\gamma) (x+\alpha)(x+\beta)(x+\gamma)$
$=-f (x) f (-x)$ (using (A))
$p (x^{2}) = (x^{3}+x+1) (x^{3}+x-1)$
$\therefore p(3^{2}) = (27+3+1) (27+3-1)$ (putting x = 3)
$= (30+1) (30-1)$
$\therefore p (9) = 899$