We know that,
(1)log_zA=P<=>A=z^P
We take ,the base of log as 10
Let ,
logx/(a^2+ab+b^2)=logy/(b^2+bc+c^2)=logz/(c^2+ca+a^2)=k
So ,
logx=k(a^2+ab+b^2)=>x=10^(k(a^2+ab+b^2))
logy=k(b^2+bc+c^2)=>y=10^(k(b^2+bc+c^2)
logz=k(c^2+ca+a^2)=>z=10^(k(c^2+ca+a^2)
Now ,
x^(a-b)=10^(k(a-b)(a^2+ab+b^2))=10^(k(a^3-b^3)
y^(b-c)=10^(k(b-c)(b^2+bc+c^2))=10^(k(b^3-c^3)
z^(c-a)=10^(k(c-a)(c^2+ca+a^2))=10^(k(c^3-a^3)
Hence ,
x^(a-b)*y^(b-c)*z^(c-a) =10^(k(a^3-b^3) )*10^(k(b^3-c^3))*10^(k(c^3-a^3)
x^(a-b)*y^(b-c)*z^(c-a) =10^((k(a^3-b^3) )+(k(b^3-c^3))+(k(c^3-a^3)
x^(a-b)*y^(b-c)*z^(c-a) =10^(k(a^3-b^3+b^3-c^3+c^3-a^3)
x^(a-b)*y^(b-c)*z^(c-a) =10^(k(0)) =10^0=1
x^(a-b)*y^(b-c)*z^(c-a) =1