if `l=1+log_a bc , m=1+log_b ca ,n=1+log_c ab` then the value of `1/l+1/m+1/n-1`

l=1+log_(a)bc,m=1+log_(b)ca,n=1+log_(c)a then the value of (1)/(l)+(1)/(m)+(1)/(n)-1

If x = log_(a) bc, y = log_(b) ca, z = log_(c) ab, then the value of (1)/(1 + x) + (1)/(1 + y) + (1)/(l + z) will be

If x=log_(a)bc,y=log_(b)ca,z=log_(c)ab, then the value of (1)/(1+x)+(1)/(1+y)+(1)/(1+z) will be

If log_(a)bc = x, log_(b)ca = y and log_(c)ab = z , then the value of (1)/(x +1) + (1)/(y +1) + (1)/(z+1) is equal to

if x =log_a (bc), y= log_b (ca) and z=log_c(ab) then 1/(x+1)+1/(y+1)+1/(z+1)

If x = 1 + log_(a) bc, y = 1 + log_(b) ca, z = 1 + log_(c) ab, then xy + yz + zx =

If x = log_(a)(bc), y = log_(b)(ca), z = log_(c)(ab) , then find 1/(x+1) + 1/(y+1) + 1/(z+1)

If =1+log_(a)bc,y=1+log_(b)ca,z=1+log_(c)ab then prove that xyz=xy+yz+zx

If x=1 + log_(a)(bc) , y=1 +log_(b)(ca) and z=1+ log_(c )(ab) prove that xy+yz+zx=xyz