Understanding the Algebraic Problem

The question provides us with two conditions involving variables \(x\), \(y\), \(z\), \(a\), \(b\), and \(c\). We are given that \(x^{2a} = y^{2b} = z^{2c}\) and that this common value is not zero. We are also given the condition \(x^2 = yz\). Our goal is to find the value of the expression \(\frac{{ab + bc + ca}}{{bc}}\).

Let's break down the problem step-by-step using the given conditions.

Step-by-Step Solution to Find the Value

The first condition is \(x^{2a} = y^{2b} = z^{2c} \ne 0\). Let's set this common value equal to a constant, say \(k\), where \(k \ne 0\). So, we have:

• \(x^{2a} = k\)
• \(y^{2b} = k\)
• \(z^{2c} = k\)

From these equations, we can express \(x\), \(y\), and \(z\) in terms of \(k\), \(a\), \(b\), and \(c\). Assuming \(a, b, c\) are such that the roots are real and defined:

• \(x = k^{\frac{1}{2a}}\)
• \(y = k^{\frac{1}{2b}}\)
• \(z = k^{\frac{1}{2c}}\)

Now, let's use the second condition given in the problem: \(x^2 = yz\).

Substitute the expressions for \(x\), \(y\), and \(z\) we found in the previous step into this equation:

\(\left(k^{\frac{1}{2a}}\right)^2 = k^{\frac{1}{2b}} \cdot k^{\frac{1}{2c}}\)

Using the exponent rules \((p^m)^n = p^{mn}\) and \(p^m \cdot p^n = p^{m+n}\), we simplify both sides of the equation:

\(k^{\frac{2}{2a}} = k^{\frac{1}{2b} + \frac{1}{2c}}\)

\(k^{\frac{1}{a}} = k^{\frac{c}{2bc} + \frac{b}{2bc}}\)

\(k^{\frac{1}{a}} = k^{\frac{c+b}{2bc}}\)

Since the bases are equal (\(k\)) and \(k \ne 0\) and we can assume \(k \ne 1\) (otherwise \(x=y=z=1\) which would satisfy \(x^2=yz\) but wouldn't constrain a, b, c meaningfully from \(x^{2a}=y^{2b}=z^{2c}=1\)), we can equate the exponents:

\(\frac{1}{a} = \frac{b+c}{2bc}\)

Now, we can cross-multiply to find a relationship between \(a\), \(b\), and \(c\):

\(1 \cdot (2bc) = a \cdot (b+c)\)

\(2bc = ab + ac\)

We are asked to find the value of the expression \(\frac{{ab + bc + ca}}{{bc}}\).

We can rewrite the expression as \(\frac{(ab + ca) + bc}{bc}\).

From our derived relationship, we know that \(ab + ac = 2bc\). Let's substitute this into the expression:

\(\frac{(2bc) + bc}{bc}\)

\(\frac{3bc}{bc}\)

Since it's given that the common value \(x^{2a} = y^{2b} = z^{2c} \ne 0\), this implies that \(x, y, z \ne \pm 1\) unless the exponents are zero, or the base is 1. If \(x,y,z\) are not zero, then \(bc\) in the denominator cannot be zero (which would imply \(k^0=1\)). Assuming \(b \ne 0\) and \(c \ne 0\), we can cancel \(bc\) from the numerator and the denominator:

\(3\)

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

Conclusion on the Value

By using the given conditions \(x^{2a} = y^{2b} = z^{2c} \ne 0\) and \(x^2 = yz\), and applying exponent rules, we derived the relationship \(2bc = ab + ac\). Substituting this into the target expression \(\frac{{ab + bc + ca}}{{bc}}\) allowed us to simplify it and find its numerical value, which is 3.

Revision Table: Exponents and Algebraic Relations

Additional Information: Related Concepts

This problem involves basic properties of exponents and algebraic manipulation. Understanding how to work with powers and roots, as well as manipulating algebraic expressions, is crucial for solving such problems. The condition \(p^m = p^n \implies m=n\) is valid when the base \(p\) is not 0, 1, or -1. If \(p=1\), any exponents would yield 1. If \(p=0\), only positive exponents matter. If \(p=-1\), the equality holds only for specific exponent values. In this problem, setting \(x^{2a} = y^{2b} = z^{2c} = k \ne 0\) and the structure of the solution implies \(k\) is usually treated as a positive value different from 1 in such contexts, allowing us to safely equate the exponents.

Problems like this test your ability to translate given conditions into equations and then solve for an unknown value or expression using algebraic techniques.