If ` x^(2a) = y^(2b) = Z^(2c) ne 0 and x^2 =yz `, then the value of `(ab+ bc + ca)/(bc)` is :

Options

A. 3
B. 3bc
C. 3ab
D. 3ac

Answer

To solve the problem, we need to find the value of \((ab + bc + ca) / (bc)\) given the conditions \(x^{2a} = y^{2b} = z^{2c} \neq 0\) and \(x^2 = yz\).

### Step 1: Set the common value
Let \(k = x^{2a} = y^{2b} = z^{2c}\). This means:
- \(x^{2a} = k\)
- \(y^{2b} = k\)
- \(z^{2c} = k\)

### Step 2: Express \(x\), \(y\), and \(z\) in terms of \(k\)
From the equations, we can express \(x\), \(y\), and \(z\) as follows:
- \(x = k^{1/(2a)}\)
- \(y = k^{1/(2b)}\)
- \(z = k^{1/(2c)}\)

### Step 3: Use the second condition \(x^2 = yz\)
Substituting the expressions for \(x\), \(y\), and \(z\) into the second condition:
\[ x^2 = (k^{1/(2a)})^2 = k^{1/a} \]
\[ yz = k^{1/(2b)} \cdot k^{1/(2c)} = k^{(1/(2b) + 1/(2c))} \]
Setting these equal gives:
\[ k^{1/a} = k^{(1/(2b) + 1/(2c))} \]

### Step 4: Equate the exponents
Since \(k \neq 0\), we can equate the exponents:
\[ \frac{1}{a} = \frac{1}{2b} + \frac{1}{2c} \]

### Step 5: Find a common denominator
To combine the right side, we find a common denominator:
\[ \frac{1}{a} = \frac{c + b}{2bc} \]
Cross-multiplying gives:
\[ 2bc = a(b + c) \]

### Step 6: Rearranging the equation
From \(2bc = a(b + c)\), we can express \(a\) in terms of \(b\) and \(c\):
\[ a = \frac{2bc}{b+c} \]

### Step 7: Substitute \(a\) into the expression \((ab + bc + ca)\)
Now we can substitute \(a\) into the expression we want to evaluate:
\[ ab + bc + ca = \left(\frac{2bc}{b+c}\right)b + bc + \left(\frac{2bc}{b+c}\right)c \]
This simplifies to:
\[ \frac{2b^2c}{b+c} + bc + \frac{2bc^2}{b+c} \]
Combining the terms gives:
\[ \frac{2b^2c + bc(b+c) + 2bc^2}{b+c} = \frac{2b^2c + b^2c + bc^2 + 2bc^2}{b+c} = \frac{3bc(b+c)}{b+c} = 3bc \]

### Step 8: Divide by \(bc\)
Now, we can find the final value:
\[ \frac{ab + bc + ca}{bc} = \frac{3bc}{bc} = 3 \]

### Final Answer
Thus, the value of \(\frac{ab + bc + ca}{bc}\) is \(3\).