We have: $b - a = \dfrac{c^2 - a^2}{b} + \dfrac{b^2 - c^2}{a} = \dfrac{ac^2 - a^3 + b^3 - bc^2}{ab} = \dfrac{(b - a)(b^2 - ab + a^2 - c^2}{ab}$ Then: $(b - a)[(b - a)^2 - c^2] = 0$ or $a = b$

parmenides51 wrote: Which triangles satisfy the equation $\frac{c^2-a^2}{b}+\frac{b^2-c^2}{a}=b-a$ when $a, b$ and $c$ are sides of a triangle? $\frac{c^2-a^2}{b}+\frac{b^2-c^2}{a}=b-a$ On simplifying we get, $ac^2-a^3+b^3-bc^2 = ab(b-a)$ $c^2(a-b) - (a^3-b^3)= -ab(a-b)$. eqn 1 Case 1:- a-b ≠0 Dividing each term in eqn 1 by a-b we get, $c^2 - (a^2+b^2+ab) = -ab$ $c^2 = a^2 + b^2$ Therefore req. Triangle is a right angled triangle Case 2:- a-b=0 $=> a=b$ Therefore req. Triangle is an isosceles triangle Two types triangles :- right angled, isosceles

parmenides51 wrote: Which triangles satisfy the equation $\frac{c^2-a^2}{b}+\frac{b^2-c^2}{a}=b-a$ when $a, b$ and $c$ are sides of a triangle? Is it a first round problem?(Because it requires just some basic algebra techniques.)

final round, not all countries have the same kind of difficulty in each olympiad

(b-a)(b²+a²-c²)=0 So, b=a, b²+a²=c² So, either isosceles or right angled triangle