Problem
ALG-B2-M06-P003 Symmetric sum of the second type
#3
★★☆☆☆ Level 2 of 5
Prove \[\sum_{\mathrm{sym}}a^2b=pq-3r.\]
Hint. Expand \((a+b+c)(ab+bc+ca)\).
Expanding \(pq\), we get the six terms \(a^2b,a^2c,b^2a,b^2c,c^2a,c^2b\) and three copies of \(abc\). Thus \(pq=\sum_{\mathrm{sym}}a^2b+3r\), which gives the formula.
Reinforces the distinction between symmetric and cyclic sums.