Question

x^2a=y^2b=z^2c and x^2=yz,xyz≠0 find ab+bc+ca/bc=?​

Answer 1

X^2a=Y^2b=Z^2c and X^2=YZ, XYZ not =0

Therefore ab+bc+ca/bc=3

Step1:

Given conditions are

X^2a=Y^2b=Z^2c and

X^2=YZ

Step 2:

Above condition will be satisfied only when X=Y=Z=1

And a=b=c=1

Step 3:

Putting the value of X, Y, Z and a, b, c on the above conditions mention in the step 1 we get

Condition 1:

1^2*1=1^2*1=1^2*1

1=1=1 [condition satisfied]

Condition 2:

1^2=1*1

1=1[conditions satisfied]

Step 4:

Now we have to calculate the value of ab+bc+ca/bc

Putting the value a=b=c=1 we get

1*1+1*1+1*1/1*1=1+1+1=3.

Answer 2

(ab+bc+ca)/bc = (x² + y² + z² )/yz

Step-by-step explanation:

x²a = y²b = z²c

x² = yz

(ab+bc+ca)/bc

= a/c  + 1  + a/b

x²a = z²c

=> a/c =  z²/x²

=> a/c =  z²/yz

=> a/c = z/y

x²a = y²b

=> a/b = y²/x²

=> a/b = y²/yz

=> a/b = y/z

a/c  + 1  + a/b

= z/y  + 1  + y/z

= (z²  + yz + y²)/yz

yz = x²

= (z²  +x² + y²)/yz

= (x² + y² + z² )/yz

(ab+bc+ca)/bc = (x² + y² + z² )/yz

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