The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$. Find $\alpha^6+\beta^6+\gamma^6+\delta^6$ The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$.
By using the substitution $y=x^3$, or by any other method, find the exact value of
$\alpha^6+\beta^6+\gamma^6+\delta^6$
This is a problem from Further Mathematics(9231) Paper 1, Question 1, 2009. I tried to solve it but was unable to figure it out, especially how to find the value of $\alpha^6+\beta^6+\gamma^6+\delta^6$. Could anyone try to solve this question and explain how they got the value?
 A: "Further mathematics" sounds like IB. So, they may suggest a solution using linear recursions.
The given polynomial
$$x^4-\color{green}{1}\cdot x^3 +\color{blue}{0}\cdot x^2 +\color{blue}{0}\cdot x - 1$$ belongs to the linear recursion
$$x_{n+4} = x_{n+3}+x_n$$
with (using Vieta)

*

*$x_0 = \alpha^0 + \beta^0 + \gamma^0 + \delta^0 = 4$

*$x_1 = \alpha + \beta + \gamma + \delta =\color{green}{1}$

*$x_2 = \alpha^2 + \beta^2 + \gamma^2 + \delta^2=(\alpha + \beta + \gamma + \delta)^2 -2\cdot \color{blue}{0} = 1$

*$x_3 = \alpha^3 + \beta^3 + \gamma^3 + \delta^3 =(\alpha + \beta + \gamma + \delta)^3-3(\alpha + \beta + \gamma + \delta)\cdot\color{blue}{0}-3\cdot\color{blue}{0} =1$
Now, since
$$\alpha^6 + \beta^6 + \gamma^6 + \delta^6 = x_6$$
you just use the recursion
$$x_4= x_3+x_0 = 1+4 = 5$$
$$x_5= x_4+x_1 = 5+1 = 6$$
$$\boxed{x_6}= x_5+x_2 = 6+1 = \boxed{7}$$
A: Define the polynomial function $p$ by
$$p(x) = x^4-x^3-1$$
Also define the function $q$ by
$q(x)=p\left(x^{1/3}\right)$. Then,
$$q(t)=t^{4/3}-t-1$$
By definition of $q$, we have that $\alpha^3,\beta^3,\gamma^3,\delta^3$ are roots of the equation $q(t)=0$.
$$q(t)=0\iff t^4 - (t+1)^3=0$$
The polynomial $t^4-(t+1)^3$ is a fourth degree polynomial in $t$, and we know four of its roots -- and since any polynomial of degree $n\geqslant 1$ has exactly $n$ roots, these are the $\textit{only}$ roots.
Now, Vieta's formulae give
$$\alpha^3+\beta^3+\gamma^3+\delta^3 = 1 \:\:\text{ and } \:\:\alpha^3\beta^3+\alpha^3\gamma^3+\alpha^3\delta^3+\beta^3\gamma^3+\beta^3\delta^3+\gamma^3\delta^3=-3$$
Thus, $$\begin{align}\alpha^6+\beta^6+\gamma^6+\delta^6&=\left(\alpha^3+\beta^3+\gamma^3+\delta^3\right)^2-2\left(\alpha^3\beta^3+\alpha^3\gamma^3+\alpha^3\delta^3+\beta^3\gamma^3+\beta^3\delta^3+\gamma^3\delta^3\right)\\ &=7\end{align}$$
The problem is solved. $\square$
A: Cubing the given equation $x^4=x^3+1$
$$x^{12} -3x^6-1=x^9+3x^3$$
and then squaring above
$$x^{24}-7x^{18}+x^{12}-3x^6+1=0$$
which is quartic in $x^6$. Thus,
$$\alpha^6+\beta^6+\gamma^6+\delta^6=7$$
A: Making the given substitution yields $y^{4/3}=y+1$, or $y^4=(y+1)^3$ or $y^4-y^3-3y^2-3y-1=0$. The roots of this polynomial are $\alpha^3,\beta^3,\gamma^3,\delta^3$, so we can work out
$$\alpha^6+\beta^6+\gamma^6+\delta^6=(\alpha^3+\beta^3+\gamma^3+\delta^3)^2-2(\alpha^3\beta^3+\cdots)=1^2-2(-3)=7$$
A: Let $P_n:=\alpha^n+\beta^n+\gamma^n+\delta^n$. Then, Newton's Sums tell us that $$1\cdot P_1-1=0\iff P_1=1\\1\cdot P_2-1\cdot P_1+2\cdot0=0\iff P_2=1\\1\cdot P_3-1\cdot P_2+0\cdot P_1+3\cdot 0=0\iff P_3=1\\\vdots$$
