Problem
ALG-B1-M01-P018 Cyclic Difference
#18
★★★★☆ Level 4 of 5
Factor \(a^2(b-c)+b^2(c-a)+c^2(a-b)\).
The expression becomes zero when \(a=b\), \(b=c\), \(c=a\), so expect factors \(a-b\), \(b-c\), \(c-a\).
Let the expression be \(E\). It has degree \(3\), and if \(a=b\), then \(E=0\), so \(a-b\) is a factor. Similarly, \(b-c\) and \(c-a\) are factors.
Since \((a-b)(b-c)(c-a)\) also has degree \(3\), only the sign remains. Substitute \(a=1\), \(b=2\), \(c=3\): \(E=1(-1)+4(2)+9(-1)=-2\). But \((a-b)(b-c)(c-a)=(-1)(-1)(2)=2\). Hence \(E=-(a-b)(b-c)(c-a)\).
A strong problem on detecting factors by zero substitution.