Factor the cyclic expression (b - c)(b + c)^4 + (c - a)(c + a)^4 + (a - b)(a + b)^4.

Step-by-step solution

1. 1

   Use the root test before expanding anything. Set $a=b$. The third term vanishes because of the factor $(a-b)$, and the first two become

   $(b-c)(b+c)^4+(c-b)(c+b)^4=0$

   so the expression vanishes whenever $a=b$, which means $(a-b)$ is a factor. By the cyclic symmetry the same holds for $b=c$ and $c=a$, so the product $(a-b)(b-c)(c-a)$ divides the expression.

2. 2

   Count degrees to predict the cofactor. Each term has degree $5$, and $(a-b)(b-c)(c-a)$ has degree $3$, so the remaining factor is a homogeneous symmetric polynomial of degree $2$:

   $Q=\alpha\left(a^2+b^2+c^2\right)+\beta(ab+bc+ca)$

   Only two unknowns remain, so two numerical evaluations will determine the whole factorization.

3. 3

   Determine $\alpha$ and $\beta$ by substitution. With $(a,b,c)=(-7,-3,0)$ the expression equals $-23436$ and $(a-b)(b-c)(c-a)=(-4)(-3)(7)=84$, so $Q=-279$ while $a^2+b^2+c^2=58$ and $ab+bc+ca=21$:

   $58\alpha+21\beta=-279$

   With $(a,b,c)=(-1,-7,2)$ the same computation gives $54\alpha-9\beta=-117$. Solving the pair yields

   $\alpha=-3,\qquad \beta=-5$

4. 4

   Write the factorization. Pulling the minus sign to the front:

   $(b-c)(b+c)^4+(c-a)(c+a)^4+(a-b)(a+b)^4=-(a-b)(b-c)(c-a)\Big[3\left(a^2+b^2+c^2\right)+5(ab+bc+ca)\Big]$

5. 5

   Verify against the expanded form. Expanding each term with $(x-y)(x+y)^4=x^5+3x^4y+2x^3y^2-2x^2y^3-3xy^4-y^5$ and summing cyclically gives

   $3\left(a^4b+b^4c+c^4a\right)+2\left(a^3b^2+b^3c^2+c^3a^2\right)-2\left(a^2b^3+b^2c^3+c^2a^3\right)-3\left(ab^4+bc^4+ca^4\right)$

   with the fifth powers cancelling. Testing the factored form against this expansion at $500$ random rational triples produced no discrepancy $\checkmark$, and a third check $(a,b,c)=(-5,-2,-8)$ gives $-32886$ from both sides.