Check for factors using the Factor Theorem.

The expression to factorize is $P(a,b,c) = a^3(b-c) + b^3(c-a) + c^3(a-b)$.
This is a cyclic expression.
If we substitute $a=b$ into the expression:

$$P(b,b,c) = b^3(b-c) + b^3(c-b) + c^3(b-b)$$ $$P(b,b,c) = b^3(b-c) - b^3(b-c) + c^3(0)$$ $$P(b,b,c) = 0$$

Since $P(b,b,c) = 0$, $(a-b)$ is a factor of $P(a,b,c)$.
By cyclic symmetry, if $(a-b)$ is a factor, then $(b-c)$ and $(c-a)$ must also be factors.
Therefore, $(a-b)(b-c)(c-a)$ is a factor of $P(a,b,c)$.

The given expression $P(a,b,c)$ is a homogeneous polynomial of degree $3+1=4$.
The product of the factors found, $(a-b)(b-c)(c-a)$, is a homogeneous polynomial of degree $1+1+1=3$.
Thus, the remaining factor must be a homogeneous polynomial of degree $4-3=1$.
Due to the cyclic symmetry of the original expression, this linear factor must be of the form $k(a+b+c)$ for some constant $k$.
So, we can write:

$$a^3(b-c) + b^3(c-a) + c^3(a-b) = k(a-b)(b-c)(c-a)(a+b+c)$$

To find $k$, we substitute specific values for $a, b, c$. Let $a=0, b=1, c=2$.
Substitute these values into the equation:
Left Hand Side (LHS):

$$0^3(1-2) + 1^3(2-0) + 2^3(0-1)$$ $$= 0 \cdot (-1) + 1 \cdot (2) + 8 \cdot (-1)$$ $$= 0 + 2 - 8$$ $$= -6$$

Right Hand Side (RHS):

$$k(0-1)(1-2)(2-0)(0+1+2)$$ $$= k(-1)(-1)(2)(3)$$ $$= k(1)(2)(3)$$ $$= 6k$$

Equating LHS and RHS:

$$6k = -6$$ $$k = -1$$

Substitute $k=-1$ back into the equation from Step 2:

$$a^3(b-c) + b^3(c-a) + c^3(a-b) = -1(a-b)(b-c)(c-a)(a+b+c)$$

This can be rewritten by absorbing the negative sign into one of the factors, for example, $(c-a)$ becomes $(a-c)$:

$$*(a-b)(b-c)(a-c)(a+b+c)*$$