Q: Use the factor theorem to show that $a + b - c$ is a factor of $f(a, b, c) = (a+b+c)^3 - 6(a + b + c)(a^2 + b^2 + c^2) + 8(a^3 + b^3 + c^3)$ and hence factorise completely.
I know that the factor theorem states for $f(x)$ if $f(a) = 0$ then $x-a$ is a factor but when there are multiple variables how do you use factor theorem?
On
Hint...if you replace each $c$ term with $(a+b)$ and expand and simplify, the entire expression is zero, so $(a+b-c)$ is a factor.
Now write down two other similar looking factors, using the fact that the whole expression is symmetrical in $a,b$ and $c$ - i.e the letters are interchangeable.
All you need to do then is to decide what is the numerical factor which will complete the factorization by comparing coefficients...
On
We are given that $$ f(a,b,c) := (a+b+c)^3 - 6(a+b+c)(a^2+b^2+c^2) + 8(a^3+b^3+c^3).$$ Define three power sum polynomials $$ p_1(a,b,c) := a+b+c, \quad p_2(a,b,c) := a^2+b^2+c^2, \quad p_3(a,b,c) := a^3+b^3+c^3. $$ Notice that $$ p_1(a,b,a+b) = 2(a+b), \qquad p_3(a,b,a+b) = (a+b)(2a^2+ab+2b^2). $$ Define a one variable polynomial $\,P(x) := f(a,b,x)\,$ and rewrite $\,P(a+b) = f(a,b,a+b)\,$ in terms of power sums to get $$ P(a+b) = 8(a+b)^3 - 12(a+b)(a^2+b^2+(a+b)^2) + 8(a+b)(2a^2+ab+2b^2) $$ which factors further as $$ P(a+b) = (a+b)\left( 8(a+b)^2 - 12(-2ab+2(a+b)^2) + 8(-3ab+2(a+b)^2)\right) $$ and simplify to get $$ P(a+b) = (a+b)\left( (8-24+16)(a+b)^2 + (24-24)ab \right) = 0. $$ Use the factor theorem to get that $\, P(x) \,$ is divisible by $\,x-(a+b).\,$ In particular, this implies that $\,P(c) = f(a,b,c)\,$ is divisible by $\,c-(a+b) = -(a+b-c).\,$
Here is a compilation of my comments. Denote $c$ by $x$, write your polynomial as a cubic polynomial in $x$ whose coefficients are polynomials in $a,b$:
$$f(x)=f_0(a,b)x^3+f_1(a,b)x^2+f_2(a,b)x+f_3(a,b).$$
Plug in $x=a+b$, you get $f(a+b)=0$. So by the theorem, $x-(a+b)$ is a factor of $f(x)$: $f(x)=(x-(a+b))g(x)$. Then your original polynomial $f(c)=(c-a-b)g(c)=(a+b-c)(-g(c))$, so $a+b-c$ is a factor.