Question

If a=b²x , b=c²y , c=a²z, find the value of xyz

Answer 1

So xyz = 1/abc.
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Answer 2

Heyy here is U r Ans :=>>>

1. solution :=>

a = b²x 
x = a/b², if b ≠ 0 

b = c²y 
y = b/c², if c ≠ 0 

c = a²z 
z = c/a², if a ≠ 0 

For a ≠ 0, b ≠ 0, c ≠ 0, 
xyz 
= (a/b²)(b/c²)(c/a²) 
= abc/(a²b²c²) 
= 1/(abc) 

If either a, b, or c is equal to zero, then all three are zero. In that case xyz may have any real value.

2. solution:=>

A = b^2 x 
divide both sides by b^2 
A / b^2 = x 

b= c^2 y 
divide both sides by c^2 
b / c^2 = y 

c = a^2 z 
divide both sides by a^2 
c / a^2 = z 

x y z = ( A / b^2) ( b / c^2) ( c / a^2) 
xyz = A b c / (a^2 b^2 c^2) 
xyz = A / (a^2 b c) 

Did you mean 'a' instead of 'A?' If so: 
xyz = a b c / (a^2 b^2 c^2) 
= 1 / (abc)

I hope this helps U :-)