JEE Polynomial 3

$\begin{aligned} P(x) & = x^6-x^5-x^3-x^2-x\\ & = x^2(x^4-x^3-x^2-1) + x^4-x^3-x \\ & = x^2(x^4-x^3-x^2-1) + (x^4-x^3-x^2-1)+x^2-x+1 \\ & = (x^2+1)(x^4-x^3-x^2-1) +x^2-x+1 \end{aligned}$

$\Rightarrow \begin{aligned} P(\alpha) & = 0 + \alpha^2-\alpha+1 \\ P(\beta) & = 0 + \beta^2-\beta+1 \\ P(\gamma) & = 0 + \gamma^2-\gamma+1 \\ P(\delta) & = 0 + \delta^2-\delta+1 \end{aligned}$

$P(\alpha) + P(\beta) + P(\gamma) + P(\delta) = \alpha^2+\beta^2 + \gamma^2 + \delta^2 -(\alpha+\beta + \gamma + \delta) +4$

Using Vieta's Formulas, we have:

$\alpha+\beta + \gamma + \delta = 1$

$\alpha\beta + \alpha\gamma + \alpha \delta+\beta\gamma + \beta\delta+ \gamma\delta = -1$

$\alpha^2+\beta^2 + \gamma^2 + \delta^2 \\ \quad = (\alpha+\beta + \gamma + \delta)^2 - 2(\alpha\beta + \alpha\gamma + \alpha \delta+\beta\gamma + \beta\delta+ \gamma\delta) \\ \quad = (1)^2-2(-1) = 3$

$\Rightarrow P(\alpha) + P(\beta) + P(\gamma) + P(\delta) = 3 -1+4 = \boxed{6}$

From the definition of P we can write

$P(\alpha)=\alpha^6-\alpha^5-\alpha^3-\alpha^2-\alpha$

Use the equation in $x$ to simplify the third and fourth terms to get

$P(\alpha)=\alpha^6-\alpha^5-\alpha^4+1-\alpha$

$\implies P(\alpha)=\alpha^2(\alpha^4-\alpha^3-\alpha^2)+1-\alpha$

Now use the equation in $x$ again to simplify the expression in brackets to get

$P(\alpha)=\alpha^2+1-\alpha$

Repeating this argument for the other roots lets us write

$P(\alpha)+P(\beta)+P(\gamma)+P(\delta)=(\alpha^2+\beta^2+\gamma^2+\delta^2)-(\alpha+\beta+\gamma+\delta)+4$

Applying Newton's identities for the sums of the powers of the roots of a polynomial to the equation in $x$ then gives the result

$P(\alpha)+P(\beta)+P(\gamma)+P(\delta)=3-1+4=\boxed{6}$

Addendum

Newton's identities are a sequence of recursive formula for finding the sums of the powers of the roots of a polynomial equation.

For the equation

$x^n+bx^{n-1}+cx^{n-2}+\dots=0$

they begin

sum of roots $= S = -b$

sum of the squares of roots $= -bS-2c$

$P(x)= x^6-x^5-x^3-x^2-x$ we only care about $x$ where $x^4-x^3-x^2-1=0$ which is when $x=\alpha, \beta, \gamma,$ or $\delta$ . So for these $x$ values $P(x)- (x^4-x^3-x^2-1)=P(x)$ .

So $P(x)=x^6-x^5-x^3-x^2-x-(x^4-x^3-x^2-1)$

$=x^6-x^5-x^4-x+1$

$=x^2(x^4-x^3-x^2)-x+1.$

since $x^4-x^3-x^2-1=0, x^4-x^3-x^2=1.$ So $P(x)= x^2-x+1$ . $P(x)-2=x^2-x-1 =\frac{x^4-x^3-x^2}{x^2}=\frac{1}{x^2}$ . Thus $P(x)=2+\frac{1}{x^2}$ .

$x^4-x^3-x^2-1=(x+1)(x^3-2x^2+x-1)$

Let $\delta=-1$ and $\alpha, \beta,$ and $\gamma$ be the roots of the equation $x^3-2x^2+x-1=0.$

$P(\alpha) + P(\beta) + P(\gamma) + P(-1) = 2+\frac{1}{\alpha^2}+2+\frac{1}{\beta^2}+2+\frac{1}{\gamma^2}+2+\frac{1}{(-1)^2}$ = $9+\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}$ .

Use vietas on $x^3-2x^2+x-1=0$ .

$\alpha+\beta +\gamma =2$

$\alpha\beta+\beta\gamma +\alpha\gamma=-1$

$\alpha\beta\gamma=1$

$\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=\frac{(\alpha\beta)^2+(\beta\gamma)^2 +(\alpha\gamma)^2}{(\alpha\beta\gamma)^2}$ = $\frac{(\alpha\beta+\beta\gamma +\alpha\gamma)^2-2(\alpha+\beta +\gamma)(\alpha\beta\gamma)}{(\alpha\beta\gamma)^2}$ =

$\frac{(-1)^2-2(2)(1)}{(1)^2}=-3$ .

$P(\alpha) + P(\beta) + P(\gamma) + P(-1) =9+\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2} =9-3=\boxed{6}$