Question

If \(\frac{x}{(b + c)(b + c - 2a)} = \frac{y}{(c - a)(c + a - 2b)} = \frac{z}{(a - b)(a + b - 2c)}\), then the value of \(x + y + z\) is:

• A. \(a + b + c\)
• B. \(a^2 + b^2 + c^2\)
• C. 0
• D. 1

Answer 1

The Correct Option is C

We are given the equation:

\(\frac{x}{(b+c)(b+c-2a)} = \frac{y}{(c-a)(c+a-2b)} = \frac{z}{(a-b)(a+b-2c)} = k\)

Here, \(x = k(b+c)(b+c-2a)\), \(y = k(c-a)(c+a-2b)\), and \(z = k(a-b)(a+b-2c)\).

To find \(x+y+z\), we sum the expressions for \(x\), \(y\), and \(z\):

\(x + y + z = k\left((b+c)(b+c-2a) + (c-a)(c+a-2b) + (a-b)(a+b-2c)\right)\)

Let's expand each term:

\((b+c)(b+c-2a) = b^2 + 2bc + c^2 - 2ab - 2ac\)

\((c-a)(c+a-2b) = c^2 - a^2 - 2bc + 2ab\)

\((a-b)(a+b-2c) = a^2 - b^2 - 2ac + 2bc\)

Summing these results:

\(b^2 + 2bc + c^2 - 2ab - 2ac + c^2 - a^2 - 2bc + 2ab + a^2 - b^2 - 2ac + 2bc\)

Combining like terms:

Observe that each term appears twice with opposite signs (e.g., \(2bc\) and \(-2bc\)), leading to a complete cancellation:

\((b^2-b^2) + (2bc-2bc) + (c^2-c^2) + (-a^2+a^2) + (-2ab+2ab) + (-2ac+2ac)\)

All terms cancel, leaving:

\(0\)

Thus, the expression inside the parentheses is zero; hence,

\(x+y+z = k \times 0 = 0\)

The value of \(x+y+z\) is 0.