What is 1/x(x-y)(x-z)+1/y(y-z)(y-x)+1/z(z-x)(z-y) equal to ?

Correct Answer is : 1xyz1/xyzxyz1

Correct Answer is : 1xyz1/xyzxyz1

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UPSC CDS 2 2022 Math Paper

Question : 10 of 100

Marks: +1, -0

What is

\frac{1}{x(x-y)(x-z)}+\frac{1}{y(y-z)(y-x)}+\frac{1}{z(z-x)(z-y)}

equal to ?

Solution:

Concept:

Use a simple substitution of numbers to evaluate the expression and match with the given options.

Explanation:

Let the given expression be

E = \frac{1}{x(x-y)(x-z)} + \frac{1}{y(y-z)(y-x)} + \frac{1}{z(z-x)(z-y)}

Substitute convenient values:

x = 1

y = 2

z = 3

Compute the first term:

\frac{1}{1(1-2)(1-3)} = \frac{1}{1 \cdot (-1) \cdot (-2)} = \frac{1}{2}

Second term:

\frac{1}{2(2-3)(2-1)} = \frac{1}{2 \cdot (-1) \cdot 1} = -\frac{1}{2}

Third term:

\frac{1}{3(3-1)(3-2)} = \frac{1}{3 \cdot 2 \cdot 1} = \frac{1}{6}

Now add:

\frac{1}{2} - \frac{1}{2} + \frac{1}{6} = \frac{1}{6}

Check the options:

Option C:

\frac{1}{xyz} = \frac{1}{1 \cdot 2 \cdot 3} = \frac{1}{6}

, which matches the computed value.

Thus the expression equals

\frac{1}{xyz}

Answer:

\frac{1}{xyz}

(Option C)

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