How Do You Solve Complex Root Expressions in Polynomial Equations?

[anemone]

Here is this week's POTW:

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Let $u,\,v,\,w$ be the roots of the equation $x^3-6x^2+18x-36=0$.

Evaluate

$\left(\dfrac{u}{v}+\dfrac{v}{u}+\dfrac{v}{w}+\dfrac{w}{v}+\dfrac{u}{w}+\dfrac{w}{u}+3\right)(3^{u^2+v^2+w^2})^{u^3+v^3+w^3}$.

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[anemone]

No one answered last week High School's POTW correctly but an honorable mention goes to lfdahl, as he computed all values correct except for the value of $\left(\dfrac{u}{v}+\dfrac{v}{u}+\dfrac{v}{w}+\dfrac{w}{v}+\dfrac{u}{w}+\dfrac{w}{u}+3\right)$.

Here's the proposed solution:

Spoiler

We're told $u,\,v,\,w$ are the roots of the equation $x^3-6x^2+18x-36=0$.

Vieta's formulas tell us then that

$u+v+w=6$; $uv+uw+vw=18$; $uvw=36$

Not that the expression inside the first parentheses could be algebraically modified so we could determine its value:

$\begin{align*}\dfrac{u}{v}+\dfrac{v}{u}+\dfrac{v}{w}+\dfrac{w}{v}+\dfrac{u}{w}+\dfrac{w}{u}+3&=1+\dfrac{u}{v}+\dfrac{u}{w}+1+\dfrac{v}{u}+\dfrac{v}{w}+1+\dfrac{w}{v}+\dfrac{w}{u}\\&=\dfrac{u}{u}+\dfrac{u}{v}+\dfrac{u}{w}+\dfrac{v}{v}+\dfrac{v}{u}+\dfrac{v}{w}+\dfrac{w}{w}+\dfrac{w}{v}+\dfrac{w}{u}\\&=u\left(\dfrac{1}{u}+\dfrac{1}{v}+\dfrac{1}{w}\right)+v\left(\dfrac{1}{u}+\dfrac{1}{v}+\dfrac{1}{w}\right)+w\left(\dfrac{1}{u}+\dfrac{1}{v}+\dfrac{1}{w}\right)\\&=(u+v+w)\left(\dfrac{1}{u}+\dfrac{1}{v}+\dfrac{1}{w}\right)\\&=(u+v+w)\left(\dfrac{uv+uw+vw}{uvw}\right)\\&=(6)\left(\dfrac{18}{36}\right)\\&=3\end{align*}$

We also have

$u^2+v^2+w^2=(u+v+w)^2-2(uv+uw+vw)=6^2-2(18)=0$ and

$u^3+v^3+w^3=(u+v+w)^3-3(u+v+w)(uv+uw+vw)+3uvw=6^3-3(6)(18)+3(36)=0$

Thus,

$\left(\dfrac{u}{v}+\dfrac{v}{u}+\dfrac{v}{w}+\dfrac{w}{v}+\dfrac{u}{w}+\dfrac{w}{u}+3\right)(3^{u^2+v^2+w^2})^{u^3+v^3+w^3}=3(3^0)^0=3^1=3$