If `(logx)/(a^2+a b+b^2)=(logy)/(b^2+b c+c^2)=(logz)/(c^2+c a+a^2),` then `x^(a-b)*y^(b-c)*z^(c-a)=`

If log x(log x)/(a^(2)+ab+b^(2))=(log y)/(b^(2)+bc+c^(2))=(log z)/(c^(2)+ca+a^(2)) then x^(a-b)*y^(b-c)*z^(c-a)=

If (logx)/(a - b) = (logy)/(b-c) = (log z)/(c -a) , then xyz is equal to :

If (logx)/(b-c)=(logy)/(c-a)=(logz)/(a-b) ,then the value of x^(b+c).y^(c+a).z^(a+b) is

If (logx)/(b-c)=(logy)/(c-a)=(logz)/(a-b) prove that (a) x^(a)y^(b)z^(c )=1 (b) x^(b+c).y^(c+a).z^(a+b)=1 (c ) x^(b^(2)+bc+c^(2) . y^(c^(2)+ca+a^(2)) . z^(a^(2)+ab+b^(2)) =1

If (log x)/(b-c) = (log y)/(c-a) = (log z)/(a-b) , then prove that x^(b^2+bc+c^2).y^(c^2+ca+a^2).z^(a^2+ab+b^2)=1

If logx/(a+b-2c)=logy/(b+c-2a)=logz/(a+c-2b) , then prove that xyz=1