Application of remainder theorem and factor theorems to multivariable polynomials. The remainder theorem and factor theorem are usually stated as follows:
The Remainder Theorem

When a polynomial p(x) is divided by x − c, the remainder is equal to the value of
p(c).

The Factor Theorem

The term x − c is a factor of a polynomial p(x) if and only if p(c)=0.

Now I am skeptic that whether these theorems are applicable on any polynomial of degree greater than or equal to 1 i.e to multi-variable polynomial  or these theorems are applicable only for single-variable polynomials because these theorems are stated for a polynomial p(x) which in my opinion refers to a single-variable polynomial.
This question came to me by the following question which I recently encountered :

Using factor theorem, show that $a - b, b-c,$ and $c-a$ are the factors of $$a(b^2-c^2) + b(c^2-a^2) + c (a^2-b^2).$$

In the above question if we try to put $a = b$, or $b = c$, or $c=a$ we will see that the value of polynomial becomes zero. It gives us some feeling that $a - b, b-c$ and $c-a$ are the factors of the given polynomial but is it correct.
Can we apply factor theorem to multivariable polynomials ?
 A: The Factor Theorem does not directly carry over to multivariate polynomials. More context and a proof of the Nullstellensatz for principal ideals are provided in this answer to Does there exist a formal statement of the Multivariable Factor Theorem?.
For the purpose of OP's example, however, the following simpler theorem suffices, which can be proved with entirely elementary means.

If $P(x_1, x_2, \dots, x_{n-1},x_n)$ and $f(x_1, x_2,\dots,x_{n-1})$ are polynomials that satisfy $$P(x_1, x_2, \dots, x_{n-1},f(x_1, x_2,\dots,x_{n-1})) \equiv 0$$ then $P$ has $x_n - f$ as a factor i.e. there exists a polynomial $Q(x_1, x_2, \dots, x_n)$ such that $$P(x_1, x_2, \dots, x_{n-1},x_n) = \left(x_n - f(x_1, x_2,\dots,x_{n-1})\right) \cdot Q(x_1, x_2, \dots, x_n)$$

Writing the polynomial as a sum of monomials $\;P(x_1, x_2, \dots,x_n) = \sum_{k_j} c_{k_1k_2\dots k_n}x_1^{k_1}x_2^{k_2}\dots x_n^{k_n}\;$ and using the identity $\;a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+\dots+b^{k-1})\;$:
$$
\begin{align}
P(x_1,x_2,\dots,x_n) &= P(x_1,x_2,\dots,x_{n-1},x_n) - P(x_1,x_2,\dots,x_{n-1},f(x_1, x_2, \dots, x_{n-1}))
\\ &= \sum_{k_j} c_{k_1k_2\dots k_n}x_1^{k_1}x_2^{k_2}\dots x_{n-1}^{k_{n-1}}\left( x_n^{k_n}-\left(f(x_1, x_2, \dots, x_{n-1})\right)^{k_n}\right)
\\ &= \left( x_n-f(x_1, x_2, \dots, x_{n-1})\right) \cdot \sum_{k_j} c_{k_1k_2\dots k_n}x_1^{k_1}x_2^{k_2}\dots x_{n-1}^{k_{n-1}}\left( x_n^{k_n-1}+ \dots\right)
\\ &= \left( x_n-f(x_1, x_2, \dots, x_{n-1})\right) \cdot Q(x_1,x_2, \dots, x_n)
\end{align}
$$
Using this result for $P(a,b,c) = a(b^2-c^2) + b(c^2-a^2) + c (a^2-b^2)$ and $f(a,b)=a$, it is easily verified that $P(a,b,f(a,b)) \equiv 0$, so $P(a,b,c)$ has a factor of $c-f(a,b)=c-a\,$. Since $P$ has cyclic symmetry, it must also have $a-b$ and $b-c$ as factors, so in the end $P(a,b,c) = \lambda (a-b)(b-c)(c-a)$ where $\lambda$ must be a constant for $P$ to have degree $\deg P = 3$, and the constant can be determined by comparing the coefficients of any of the monomials, which gives $\lambda = 1$.
A: You can apply factor theorem to multivariable polynomials as follows-
Here you have a multivariable polynomial
$P(a,b,c)=a(b^2-c^2) + b(c^2-a^2) + c (a^2-b^2)$
Now if you find some polynomial $g(b,c)$ such that
$P(g(b,c),b,c)=0$ for all values of $b$ and $c$, then you may say that $(a-g(b,c))$ is a factor of above polynomial.
This is factor theorem for polynomials you are interested in.
For example, Here you found a polynomial $g(b,c)=b+0.c$. which when replaced by $a$ in original polynomial. then it evaluated to $0$, irrespective of what $b,c$ are, Hence $a-(g(b,c))=a-b$ is a factor of above polynomial.
Now suppose you divide this polynomial by some $a-g(b,c)=a-(b+c)$ where $g(b,c)=b+c$, then the remainder you will get will be $P(g(b,c),b,c)=-b^2c+c^2b$ which you can check by long division (as polynomials in a).
also when we had a single variable polynomial we wrote as
$P(x)= d(x).Q(x)+R(x)$. Here you can write
$P(a,b,c)=(a-g(b,c)).(a(c-b)+c^2b-b^2c$ where our divisor was $d(a,b,c)=a-g(b,c)$, quotient which we obtained by long division was $Q(a,b,c)=a(c-b)$ and remainder as
$R(a,b,c)=0.a-b^2c+c^2b$
In nutshell you can treat multivariate polynomials as single-variable polynomials only as @dxiv explained in his comments
