If $\frac{a}{b}+\frac{b}{a}=1,a\ne 0,b\ne 0,$ then find the value of $\frac{a^3+b^3}{3ab}.$

The correct answer is 0.

Let us solve the problem step-by-step.

We are given the algebraic equation:
$\frac{a}{b}+\frac{b}{a}=1$
where $a\ne 0$ and $b\ne 0$.

First, we simplify the given equation by finding a common denominator for the terms on the left-hand side:
$\frac{a^2+b^2}{ab}=1$

Multiplying both sides of the equation by $ab$ (since $a\ne 0$ and $b\ne 0$), we get:
$a^2+b^2=ab$

Rearranging the terms to set the equation to zero:
$a^2-ab+b^2=0$

Next, we recall the algebraic identity for the sum of cubes:
$a^3+b^3=(a+b)(a^2-ab+b^2)$

Substituting $a^2-ab+b^2=0$ into the identity:
$a^3+b^3=(a+b)\times 0=0$

We need to find the value of the expression:
$\frac{a^3+b^3}{3ab}$

Substituting $a^3+b^3=0$ into this expression:
$\frac{0}{3ab}=0$

Thus, the value of the expression is 0.