If one root of the equation $ax^{2}+bx+c=0$ is equal to the $n^{th}$ power of the other, then $\left ( ac^{n} \right )^{1/n+1}+\left ( a^{n}c \right )^{1/n+1}+b$ is equal to

Let the roots of the equation ax^2 + bx + c = 0 be r and r^n; by Vieta's formulas, their sum r plus r^n equals negative b divided by a, and their product r raised to the power of (n+1) equals c divided by a. Raising the sum relation to the (n+1)th power gives (r plus r^n)^(n+1) equals (negative b divided by a) raised to the (n+1)th power. We factor out r^n from the left side to get r^n times (1 plus r^(n-1))^(n+1). This is equivalent to testing the standard algebraic identity for this specific root relationship, which states that (ac^n)^(1/(n+1)) plus (a^n c)^(1/(n+1)) plus b equals 0. The result is 0.

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