MHB Solution for Roots of x^3-7x+7=0: Week #340 Nov 13th, 2018 POTW

[anemone]

Here is this week's POTW:

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Given that $a,\,b$ and $c$ are roots for the equation $x^3-7x+7=0$.

Evaluate $\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}$.

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[anemone]

Congratulations to the following members for their correct solution:

1. castor28
2. Olinguito
3. lfdahl
4. kaliprasad

Solution from castor28:

Spoiler

Let us write $u=\dfrac1a$, $v = \dfrac1b$, $w = \dfrac1c$, $t(n) = u^n + v^n + w^n$. We want to find $t(4)$.

$u$, $v$, and $w$ are roots of the reciprocal equation $7x^3 - 7x^2 + 1 = 0$. $u^n$, $v^n$, $w^n$ and therefore $t(n)$ satisfy the corresponding linear recurrence relation $7x_n= 7x_{n-1} - x_{n-3}$.

We have $t(0) = 3$ (obvious), $t(1) = 1$ (by Viete's relations). We compute:
\begin{align*}
t(2) &= u^2 + v^2 + w^2\\
&= (u+v+w)^2 - 2(uv+uw+vw)\\
&= 1^2 - 2\cdot0\\
&= 1
\end{align*}

We can now use the recurrence to compute:
\begin{align*}
t(3) &= t(2) - \frac{t(0)}{7} = 1 - \frac37 = \frac47\\
t(4) &= t(3) - \frac{t(1)}{7} = \frac47 - \frac17 = {\bf \frac37}
\end{align*}

Alternative solution from Olinguito:

Spoiler

$a,b,c$ are roots of $x^3-7x+7=x(x^2-7)+7=0$
​

$\implies\ a^2,b^2,c^2$ are roots of $\sqrt x(x-7)+7=0$

$\implies$ they are roots of $x(x-7)^2=49$

$\implies$ they are roots of $x^3-14x^2+49x-49=0$

$\implies$ they are roots of $x(x^2+49)=14x^2+49$

$\implies\ a^4,b^4,c^4$ are roots of $\sqrt x(x+49)=14x+49$

$\implies$ they are roots of $x(x+49)^2=(14x+49)^2$

$\implies$ they are roots of $x^3-98x^2+1029x-2401=0$

$\implies\ \dfrac1{a^4},\dfrac1{b^4},\dfrac1{c^4}$ are roots of $\dfrac1{x^3}-\dfrac{98}{x^2}+\dfrac{1029}x-2401=0$

$\implies$ they are roots of $2401x^3-1029x^2+98x-1=0$

$\implies \dfrac1{a^4}+\dfrac1{b^4}+\dfrac1{c^4}=\dfrac{1029}{2401}=\boxed{\dfrac37}$.