ta có
\(\text{2ap + bc = 2p.a + bc = (a + b + c).a + bc }\\ =a^2+ab+ca+bc\\ =a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
tương tự ta có
\(\left\{{}\begin{matrix}2pb+ac=\left(b+a\right)\left(b+c\right)\\2pc+ab=\left(c+a\right)\left(c+b\right)\end{matrix}\right.\)
\(\Rightarrow\left(2pa+bc\right)\left(2pb+ac\right)\left(2pc+ab\right)=\left(a+b\right)\left(c+a\right)\left(a+b\right)\left(b+c\right)\left(c+a\right)\left(b+c\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)