Power = Weak?

$\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{c}{a}+3$

$= 1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1$

$=a\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+b\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)$

$=(a+b+c)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)$

$=(a+b+c)\left(\dfrac{ab+bc+ac}{abc}\right)$

By Veita's relation, substituting the values,

$=(6)\left(\dfrac{18}{36}\right) = 3$

We also have $a^2+b^2+c^2 = (a+b+c)^2 -2(ab+bc+ac) = 6^2-2(18)= 0$

We also have,

$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ac)+3abc \\= 6^3-3(6)(18)+3(36)=0$

Thus the given expression $= \large 3(3^{(0)(0)})=3^1=\huge\boxed{\color{#D61F06}{3}}$

Use Newton's identities for working out the sums of powers of a polynomial equation to find

$a+b+c=--6=6\dots(1)$

$a^2+b^2+c^2=--6\times 6-2\times 18=0\dots(2)$

$a^3+b^3+c^3=--6\times 0-18\times 6-3\times -36=0\dots(3)$

Since both exponents are now seen to be zero, the expression simplifies to

$\dfrac{a^2c+b^2+c+b^2a+c^2a+a^2b+c^2b}{abc}+3$

$=\dfrac{a^2(c+b)+b^2(a+c)+c^2(a+b)}{abc}+3$

Using (1) this becomes

$=\dfrac{a^2(6-a)+b^2(6-b)+c^2(6-c)}{abc}+3$

$=\dfrac{6(a^2+b^2+c^2)-(a^3+b^3+c^3)}{abc}+3$

Using (2) and (3) again the expression simplifies to $\boxed{3}$

The factorisation of the equation is $(x - a)(x - b)(x - c) = 0.$ From this, we deduce that $a + b + c = 6 \\ ab + bc + ca = 18 \\ abc = 36$ from which we derive $(a + b + c)^2 = 36 \\ \Rightarrow a^2 + b^2 + c^2 + 2(ab + bc + ca) = 36 \\ \Rightarrow a^2 + b^2 + c^2 + 36 = 36 \\ \Rightarrow a^2 + b^2 + c^2 = 0 \\ \Rightarrow 3^{a^2 + b^2 + c^2} = 1 \\ \Rightarrow (3^{a^2 + b^2 + c^2}) ^{a^3 + b^3 + c^3} = 1.$ This reduces the problem to evaluating $\frac{a}{b} + \frac{b}{a} + \frac{b}{c} + \frac{c}{b} + \frac{a}{c} + \frac{c}{a} + 3 \\ \\ = \frac{a^2c + b^2c + ab^2 + ac^2 + a^2b + bc^2 + 3abc}{abc} \\ \\ = \frac{(a + b + c)(ac + bc + ca)}{abc} \\ \\ = \frac{6 \times 18}{36} \\ \\ = \boxed{3}.$