Problem
ALG-B1-M02-P024 Fifth Powers with Zero Sum
#24
★★★★★ Level 5 of 5
Let \(a+b+c=0\). Prove that \(a^5+b^5+c^5=\frac{5}{2}abc(a^2+b^2+c^2)\).
Use \(c=-(a+b)\) or a recurrence for power sums when \(a+b+c=0\).
Let \(s_k=a^k+b^k+c^k\). Since \(a,b,c\) are roots of \(t^3+qt-r=0\), where \(q=ab+bc+ca\), \(r=abc\), each root satisfies \(t^3=-qt+r\).
Multiplying by \(t^2\) and summing over the three roots gives \(s_5=-q s_3+r s_2\).
When \(a+b+c=0\), we have \(s_3=3abc=3r\), and \(s_2=-2q\). Thus \(s_5=-3qr-2qr=-5qr\).
Since \(q=-\frac{s_2}{2}\), we obtain \(s_5=\frac{5}{2}rs_2=\frac{5}{2}abc(a^2+b^2+c^2)\).
A strong problem: the student must keep the structure, but without heavy book notation.