Analyzing the Algebraic Problem

The problem provides an equation relating three variables, \(x\), \(y\), and \(z\): \(x^2 + y^2 + z^2 = xy + yz + zx\). We are also given that \(x = 1\). The goal is to evaluate a specific algebraic expression involving \(x\), \(y\), and \(z\).

Understanding the Key Condition: \(x^2 + y^2 + z^2 = xy + yz + zx\)

The given equation \(x^2 + y^2 + z^2 = xy + yz + zx\) is a standard mathematical identity in disguise. Let's rearrange and manipulate this equation to understand what it implies about the relationship between \(x\), \(y\), and \(z\).

We can multiply the entire equation by 2:

\(2(x^2 + y^2 + z^2) = 2(xy + yz + zx)\)

\(2x^2 + 2y^2 + 2z^2 = 2xy + 2yz + 2zx\)

Now, move all terms to one side of the equation:

\(2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx = 0\)

We can group the terms to form perfect squares:

\((x^2 - 2xy + y^2) + (y^2 - 2yz + z^2) + (z^2 - 2zx + x^2) = 0\)

This simplifies to:

\((x-y)^2 + (y-z)^2 + (z-x)^2 = 0\)

Since the square of any real number is non-negative (greater than or equal to zero), the only way the sum of three squares can be zero is if each individual square term is zero. This means:

• \((x-y)^2 = 0 \implies x - y = 0 \implies x = y\)
• \((y-z)^2 = 0 \implies y - z = 0 \implies y = z\)
• \((z-x)^2 = 0 \implies z - x = 0 \implies z = x\)

Therefore, the condition \(x^2 + y^2 + z^2 = xy + yz + zx\) is true if and only if \(x = y = z\).

Solving the Expression with \(x = y = z\) and \(x = 1\)

We are given that \(x = 1\) and we have deduced that \(x = y = z\). Combining these, we get \(x = y = z = 1\).

Now we need to substitute \(x = 1\), \(y = 1\), and \(z = 1\) into the given expression:

\(\rm \frac{10x^4+5y^4+7z^4}{13x^2y^2+6y^2z^2+3z^2x^2}\)

Step-by-Step Calculation

Substitute the values \(x=1\), \(y=1\), and \(z=1\) into the numerator and the denominator separately.

Numerator:

\(10x^4 + 5y^4 + 7z^4\)

\(= 10(1)^4 + 5(1)^4 + 7(1)^4\)

\(= 10(1) + 5(1) + 7(1)\)

\(= 10 + 5 + 7\)

\(= 22\)

Denominator:

\(13x^2y^2 + 6y^2z^2 + 3z^2x^2\)

\(= 13(1)^2(1)^2 + 6(1)^2(1)^2 + 3(1)^2(1)^2\)

\(= 13(1)(1) + 6(1)(1) + 3(1)(1)\)

\(= 13 + 6 + 3\)

\(= 22\)

Now, divide the numerator by the denominator:

\(\rm \frac{22}{22} = 1\)

Final Result

The value of the expression \(\rm \frac{10x^4+5y^4+7z^4}{13x^2y^2+6y^2z^2+3z^2x^2}\) under the given conditions \(x^2 + y^2 + z^2 = xy + yz + zx\) and \(x = 1\) is 1.

Revision Table: Key Concepts

Additional Information: Variations of the Condition

The condition \(x^2 + y^2 + z^2 = xy + yz + zx\) is a classic result. It often appears in different forms. For example:

• If \(x\), \(y\), and \(z\) are sides of a triangle, and \(x^2 + y^2 + z^2 = xy + yz + zx\), then the triangle must be equilateral (since \(x=y=z\)).
• The expression \(x^2 + y^2 + z^2 - xy - yz - zx\) is always non-negative for real numbers \(x\), \(y\), and \(z\). It is equal to \(\frac{1}{2}[(x-y)^2 + (y-z)^2 + (z-x)^2]\).