Playing hide and seek with $abc$

Algebra Level 3

Let $a,b,c$ positive real numbers so that: $a(b+c)=152$ $b(a+c)=162$ $c(a+b)=170$

Find the value of $abc$ .

1 solution

Paola Ramírez
Jul 21, 2015

$a(b+c)+b(a+c)+c(a+b)=2ab+2bc+2ac=484 \Rightarrow ab+bc+ac=242$

$ab+bc+ac-a(b+c)=bc=242-152=90$

$ab+bc+ac-b(a+c)=ac=242-162=80$

$ab+bc+ac-c(a+b)=ab=242-170=72$

$\Rightarrow ab\times ac\times bc=(abc)^2=90\times80\times 72 \therefore abc=\sqrt{90\times80\times72}=\boxed{720}$