Factor a²b²(a-b) +b²c²(b-c) +c²a²(c-a)

Question

This is the question that my friend asked me I tried a lot but failed.I am not finding any identity useful

Answer 1

Note that if we set $a = b$, the entire expression becomes $0$, so by the factor theorem $a-b$ is a factor, and we similarly obtain $b-c$ and $c-a$ as factors. So we wish to write $a^3b^2 - a^2b^3 + b^2c^3 - b^2c^3 + c^3a^2 - c^2a^3 = (a-b)(b-c)(c-a)Q(a,b,c)$ for some polynomial $Q$, and $Q$.

Note that the coefficient of $a^3$ on the LHS is $b^2 - c^2$, and on the RHS it is $-(b-c)Q_a(b,c)$, where $Q_a$ is the coefficient of $a$ in $Q$. So we must have $-ab - ac$ in $Q$, and by symmetry $-ab - bc - ca$ appears in $Q$. By inspection we note that this is the exact factorisation, so the answer is $-(a-b)(b-c)(c-a)(ab+bc+ca).$

Answer 2

You can expand the products to achieve this:

$a^3b^2-a^2b^3+b^3c^2-b^2c^3+c^3a^2-c^2a^3$

$a^3(b^2-c^2)+b^3(c^2-a^2)+c^3(a^2-b^2)$

Do you want to reach something smaller?