Step 1: Set up notation using Vieta formulas.
Let the roots of $ax^2+bx+c=0$ be $\alpha$ and $\alpha^n$ (one root is the $n$th power of the other). By Vieta: sum of roots $= \alpha+\alpha^n = -b/a$, and product of roots $= \alpha \cdot \alpha^n = \alpha^{n+1} = c/a$.
Step 2: Express $\alpha$ explicitly from the product relation.
From $\alpha^{n+1} = c/a$, we get $\alpha = (c/a)^{1/(n+1)}$. And $\alpha^n = (c/a)^{n/(n+1)}$.
Step 3: Multiply the sum relation through by $a$.
$\alpha+\alpha^n = -b/a$ implies $a\alpha + a\alpha^n = -b$. This is the key equation we will simplify.
Step 4: Simplify $a\alpha$ using the product relation.
$a\alpha = a \cdot (c/a)^{1/(n+1)} = a^{1-1/(n+1)} \cdot c^{1/(n+1)} = a^{n/(n+1)} c^{1/(n+1)} = (a^n c)^{1/(n+1)}$.
Step 5: Simplify $a\alpha^n$ similarly.
$a\alpha^n = a \cdot (c/a)^{n/(n+1)} = a^{1-n/(n+1)} \cdot c^{n/(n+1)} = a^{1/(n+1)} c^{n/(n+1)} = (ac^n)^{1/(n+1)}$.
Step 6: Combine and state the final answer.
From Step 3: $(a^n c)^{1/(n+1)} + (ac^n)^{1/(n+1)} = -b$. \[ \boxed{(ac^n)^{\frac{1}{n+1}}+(a^nc)^{\frac{1}{n+1}} = -b} \]