If the equation $$a(b-c)x^{2}+b(c-a)x+c(a-b)=0$$ has equal roots, where a+c=15 and $$b=\frac{36}{5}$$, then $$a^{2}+c^{2}$$ is equal to

We are given the equation $$a(b-c)x^2 + b(c-a)x + c(a-b) = 0$$ with equal roots, $$a + c = 15$$, and $$b = \dfrac{36}{5}$$.

Substituting $$x = 1$$:

$$a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ab + ac - bc = 0$$

So $$x = 1$$ is always a root.

By Vieta's formulas, the product of roots:

$$1 \times 1 = \frac{c(a-b)}{a(b-c)}$$ $$a(b - c) = c(a - b)$$ $$ab - ac = ca - cb$$ $$ab + bc = 2ac$$ $$b(a + c) = 2ac$$

$$b(a + c) = 2ac$$:

$$\frac{36}{5} \times 15 = 2ac$$ $$\frac{540}{5} = 2ac$$ $$108 = 2ac$$ $$ac = 54$$ $$a^2 + c^2 = (a + c)^2 - 2ac = 15^2 - 2(54) = 225 - 108 = 117$$

The answer is $$\boxed{117}$$.