Question:
If
\[
y^2+z^2=ayz,\quad z^2+x^2=bxz,\quad x^2+y^2=cxy,
\]
then the value of
\[
\frac{xz}{y^2}+\frac{y^2}{zx}
\]
is:
If
\[
y^2+z^2=ayz,\quad z^2+x^2=bxz,\quad x^2+y^2=cxy,
\]
then the value of
\[
\frac{xz}{y^2}+\frac{y^2}{zx}
\]
is:
Show Hint
In symmetric algebraic equations, divide each equation by the product of the two variables present on the right-hand side. This often creates useful reciprocal sums.
Updated On: Jun 18, 2026
- \(a^2-b^2+c^2\)
- \(a^2+b^2-c^2\)
- \(ac-b\)
- \(ab-c\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Divide the first equation by \(yz\).
Given, \[ y^2+z^2=ayz. \] Dividing by \(yz\), we get \[ \frac{y}{z}+\frac{z}{y}=a. \]
Step 2: Divide the third equation by \(xy\).
Given, \[ x^2+y^2=cxy. \] Dividing by \(xy\), we get \[ \frac{x}{y}+\frac{y}{x}=c. \]
Step 3: Divide the second equation by \(xz\).
Given, \[ z^2+x^2=bxz. \] Dividing by \(xz\), we get \[ \frac{z}{x}+\frac{x}{z}=b. \]
Step 4: Multiply the expressions for \(a\) and \(c\).
Now, \[ ac= \left(\frac{y}{z}+\frac{z}{y}\right) \left(\frac{x}{y}+\frac{y}{x}\right). \] Expanding, \[ ac= \frac{x}{z}+\frac{y^2}{xz}+\frac{xz}{y^2}+\frac{z}{x}. \] Rearranging, \[ ac= \left(\frac{x}{z}+\frac{z}{x}\right) + \left(\frac{xz}{y^2}+\frac{y^2}{xz}\right). \] Since \[ \frac{x}{z}+\frac{z}{x}=b, \] we get \[ ac=b+\left(\frac{xz}{y^2}+\frac{y^2}{xz}\right). \] Therefore, \[ \frac{xz}{y^2}+\frac{y^2}{xz}=ac-b. \]
Step 5: Final conclusion.
Hence, \[ \boxed{ac-b} \]
Given, \[ y^2+z^2=ayz. \] Dividing by \(yz\), we get \[ \frac{y}{z}+\frac{z}{y}=a. \]
Step 2: Divide the third equation by \(xy\).
Given, \[ x^2+y^2=cxy. \] Dividing by \(xy\), we get \[ \frac{x}{y}+\frac{y}{x}=c. \]
Step 3: Divide the second equation by \(xz\).
Given, \[ z^2+x^2=bxz. \] Dividing by \(xz\), we get \[ \frac{z}{x}+\frac{x}{z}=b. \]
Step 4: Multiply the expressions for \(a\) and \(c\).
Now, \[ ac= \left(\frac{y}{z}+\frac{z}{y}\right) \left(\frac{x}{y}+\frac{y}{x}\right). \] Expanding, \[ ac= \frac{x}{z}+\frac{y^2}{xz}+\frac{xz}{y^2}+\frac{z}{x}. \] Rearranging, \[ ac= \left(\frac{x}{z}+\frac{z}{x}\right) + \left(\frac{xz}{y^2}+\frac{y^2}{xz}\right). \] Since \[ \frac{x}{z}+\frac{z}{x}=b, \] we get \[ ac=b+\left(\frac{xz}{y^2}+\frac{y^2}{xz}\right). \] Therefore, \[ \frac{xz}{y^2}+\frac{y^2}{xz}=ac-b. \]
Step 5: Final conclusion.
Hence, \[ \boxed{ac-b} \]
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