Giải thích các bước giải:
Ta có :
$\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}$
$\to \dfrac{ab-bc}{a+b-(b+c)}=\dfrac{bc-ca}{b+c-(c+a)}=\dfrac{ca-ab}{c+a-(a+b)}$
$\to \dfrac{b(a-c)}{a-c}=\dfrac{c(b-a)}{b-a}=\dfrac{a(c-b)}{c-b}$
$\to b=c=a$
$\to P=\dfrac{(a.a+a.a+a.a)^{1008}}{a^{2016}+a^{2016}+a^{2016}}$
$\to P=\dfrac{(3a^2)^{1008}}{3a^{2016}}$
$\to P=\dfrac{3^{1008}.a^{2016}}{3a^{2016}}$
$\to P=3^{1007}$
