Rewrite the given equality in the following way:
$$\frac{a((a+b)^2-c^2)}{(a+b)^2} = \frac{c((b+c)^2-a^2)}{(b+c)^2} \quad \text{or} \quad \frac{a(a+b-c)(a+b+c)}{(a+b)^2} = \frac{c(b+c-a)(b+c+a)}{(b+c)^2}.$$
After dividing by $a+b+c$ we obtain
$$\frac{a(a+b-c)}{(a+b)^2} = \frac{c(b+c-a)}{(b+c)^2}.$$
Suppose that $c \ge a+b$. Then $b+c > a$, so the left side of the equality is not positive while the right side is positive. This contradiction means that $b+c > a$, $a+c > b$ and $b+a > c$ so we can treat the sides $a, b, c$ as sides of a triangle.
But the given equalities literally mean that the angle bisectors of this triangle are equal. Indeed: $l_c^2 = ab - a_1b_1$, where $a_1, b_1$ are the lengths of the two parts of the opposite side which are derived after drawing the angle bisector. Then we have: $a_1 + b_1 = c$ and $\frac{a_1}{b_1} = \frac{a}{b}$. It is easy to see that $a_1 = \frac{ac}{a+b}$ and $b_1 = \frac{bc}{a+b}$. Then $l_c^2 = ab - a_1b_1 = ab - \frac{abc^2}{(a+b)^2} = ab\left(1 - \frac{c^2}{(a+b)^2}\right)$. Since if three angle bisectors are equal the triangle is regular, we obtain that the solutions are $(t, t, t)$, where $t > 0$.