设两根为x1,x2
∴x1+x2=-b/a=p
x1x2=c/a
q=x1²+x2²=(x1+x2)²-2x1x2=(b²-2ac)/a²
r=x1³+x2³=(x1+x2)[(x1+x2)²-3x1x2]=-b/a(b²/a²-3c/a)
∴ar+bq+cp
=-b(b²-3ac)/a²+b(b²-2ac)/a²-bc/a
=[-b³+3abc+b³-2abc-abc]/a²
=0