We know that,
r1=△s−a, r2=△s−b, r3=△s−c
Now, (1r1+1r2)(1r2+1r3)(1r3+1r1)=(s−a△+s−b△)(s−b△+s−c△)(s−c△+s−a△)=(2s−(a+b)△)(2s−(b+c)△)(2s−(a+c)△)=c△⋅a△⋅b△=abc△3 …(1)
∵R=abc4△⇒△=abc4R
Putting the value of △ in equation (1),
abc(abc)3×(4R)3=64R3a2b2c2=KR3a2b2c2
⇒K=64