I have a really hard question from IMO (1984) it goes:
Find the factors of a^3(b-c) + b^3(c-a) + c^3(a-b). Does anyone know how to solve this? If possible, show working/reasoning as well? Thanks
Hello, cedricc!
We will factor "by grouping" ... repeatedly.
Factor: .$\displaystyle a^3(b-c) + b^3(c-a) + c^3(a-b)$
We have: .$\displaystyle a^3b - a^3c + b^3c - ab^3 + ac^3 - bc^3$
Rearrange: .$\displaystyle a^3b - a^3c - ab^3+ac^3 + b^3c-bc^3$
Factor: .$\displaystyle a^3(b-c) - a(b^3-c^3) + bc(b^2-c^2) $
Factor: .$\displaystyle a^3(b-c) - a(b-c)(b^2 + bc + c^2) + bc(b-c)(b+c)$
Factor: .$\displaystyle (b-c)\bigg[a^3 - a(b^2+bc+c^2) + bc(b+c)\bigg]$
Simplify: .$\displaystyle (b-c)\bigg[a^3 - ab^2 - abc - ac^2 + b^2c + bc^2\bigg]$
Rearrange: .$\displaystyle (b-c)\bigg[b^2c - ab^2 + bc^2 - abc - ac^2 + a^3\bigg]$
Factor: .$\displaystyle (b-c)\bigg[b^2(c-a) + bc(c-a) - a(c^2-a^2)\bigg]$
Factor: .$\displaystyle (b-c)\bigg[b^2(c-a) + bc(c-a) - a(c-a)(c+a)\bigg] $
Factor: .$\displaystyle (b-c)(c-a)\bigg[b^2 + bc - a(c+a)\bigg]$
Simplify: .$\displaystyle (b-c)(c-a)\bigg[b^2 + bc - ac - a^2\bigg]$
Rearrange: .$\displaystyle (b-c)(c-a)\bigg[b^2-a^2 + bc - ac\bigg]$
Factor: .$\displaystyle (b-c)(c-a)\bigg[(b-a)(b+a) + c(b-a)\bigg] $
Factor: .$\displaystyle (b-c)(c-a)(b-a)\bigg[b + a + c\bigg]$
Therefore: .$\displaystyle (a-b)(b-c)(a-c)(a+b+c)$