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?
We are given the equation:
\[
\frac{c^2 - a^2}{b} + \frac{b^2 - c^2}{a} = b - a
\]
where \(a\), \(b\), and \(c\) are the sides of a triangle. We need to determine which types of triangles satisfy this equation.

1. **Simplify the given equation:**
   \[
   \frac{c^2 - a^2}{b} + \frac{b^2 - c^2}{a} = b - a
   \]
   Combine the fractions on the left-hand side:
   \[
   \frac{a(c^2 - a^2) + b(b^2 - c^2)}{ab} = b - a
   \]
   Simplify the numerator:
   \[
   \frac{ac^2 - a^3 + b^3 - bc^2}{ab} = b - a
   \]
   Multiply both sides by \(ab\) to clear the denominator:
   \[
   ac^2 - a^3 + b^3 - bc^2 = ab(b - a)
   \]
   Simplify the right-hand side:
   \[
   ac^2 - a^3 + b^3 - bc^2 = ab^2 - a^2b
   \]

2. **Rearrange the equation:**
   \[
   ac^2 - bc^2 - a^3 + b^3 = ab^2 - a^2b
   \]
   Factor out common terms:
   \[
   c^2(a - b) - (a^3 - b^3) = ab(b - a)
   \]
   Notice that \(a^3 - b^3\) can be factored as \((a - b)(a^2 + ab + b^2)\):
   \[
   c^2(a - b) - (a - b)(a^2 + ab + b^2) = -ab(a - b)
   \]
   Factor out \((a - b)\) from the left-hand side:
   \[
   (a - b)(c^2 - a^2 - ab - b^2) = -ab(a - b)
   \]

3. **Case 1: \(a \neq b\):**
   Divide both sides by \((a - b)\):
   \[
   c^2 - a^2 - ab - b^2 = -ab
   \]
   Simplify:
   \[
   c^2 = a^2 + b^2
   \]
   This is the Pythagorean theorem, which implies that the triangle is a right-angled triangle.

4. **Case 2: \(a = b\):**
   Substitute \(a = b\) into the original equation:
   \[
   \frac{c^2 - a^2}{a} + \frac{a^2 - c^2}{a} = a - a
   \]
   Simplify:
   \[
   \frac{c^2 - a^2 + a^2 - c^2}{a} = 0
   \]
   \[
   \frac{0}{a} = 0
   \]
   This is always true, so the triangle is an isosceles triangle with \(a = b\).

Conclusion:
The triangles that satisfy the given equation are either right-angled triangles or isosceles triangles.

The final answer is \(\boxed{\text{right-angled or isosceles triangles}}\).