How to factorize expression $- a^{2} b + a^{2} c + a b^{2} - a c^{2} - b^{2} c + b c^{2}$?

Question

I have an expression and I know its factorized form: $$ - a^{2} b + a^{2} c + a b^{2} - a c^{2} - b^{2} c + b c^{2}=(c-a)(a-b)(b-c) $$ It's simple algebra, but I'm don't see any way how to factorize the expression in the left side by myself. Seems I missed some algebra lessons :)

I tried factorizing $a,b,c$ terms, but I don't see any way to factorize it further: $$ a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)=a(b-c)(b+c)+b(c-a)(c+a)+c(a-b)(a-b) $$

Could you help me understand how to think to factorize the expression by myself?

Answer 1

Here it is one possible approach:

\begin{align*} -a^{2}b + a^{2}c + ab^{2} - ac^{2} - b^{2}c + bc^{2} & = (ab^{2} - a^{2}b) + (a^{2}c - ac^{2}) + (bc^{2} - b^{2}c)\\\\ & = ab(b - a) + ac(a - c) + bc(c - b)\\\\ & = ab(b - a) + ac((a - b) + (b - c)) + bc(c - b)\\\\ & = (ac - ab)(a - b) + (ac - bc)(b - c)\\\\ & = a(c - b)(a - b) + c(a - b)(b - c)\\\\ & = (a - b)(b - c)(c - a) \end{align*}

Answer 2

Well, notice that the degrees in the terms on the left hand side count up to three. So, something of the form $(a+b+c)^3$ will be a factorization of it (if it factorizes).

You can also see that the left hand side is symmetric in the sense that if you permute $a,b,c$, you will get the same left hand side (there six permutations on these three letters). Therefore, your factorization should be symmetric as well.

Furthermore, you could try to fill in some simple substitutions, like $a = 2b$ (or $a =b$), to find a zero of the polynomial.

Answer 3

Rewrite the polynomial as follows :

$$ \begin{align}a^2(c-b)-a(c^2-b^2)+bc(c-b) \end{align} $$

Since $c-b$ is a common factor, you have :

$$(c-b)\left(a^2-a(b+c)+bc\right)$$

Then recall that Vieta's formulas for quadratic polynomial :

$$ \begin{align}&a^2-a(b+c)+bc=0\\ \implies &a_1=b,\quad a_2=c\end{align} $$

Finally, using by factor theorem , you get :

$$(c-b)(a-b)(a-c)$$

which completes the answer .

Answer 4

The thing to notice is that when you replace all the $b$s with $a$s, the expression is zero.

Likewise, replacing $b$ with $c$ and $c$ with $a$, the expression is zero.

So by the Factor Theorem, $(a-b)$, $(b-c)$ and $(c-a)$ are all factors.

Since the polynomial is of order $3$, there are no other factors other than these containing $a,b$ and $c$, so it factorises as $$\lambda(a-b)(b-c)(c-a)$$

Then the choice of $\lambda$ is obvious.