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UPSC CDS I 2025 Mathematics Paper

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Question : 36 of 100

Marks: +1, -0

What is

\frac{(a-b)^{2}}{(b-c)(c-a)}+\frac{(b-c)^{2}}{(c-a)(a-b)}+\frac{(c-a)^{2}}{(a-b)(b-c)}

- 3 equal to, where

≠

?

0
3
a + b + c
3 (a−b) (b−c) (c−a)

Solution:

Concept:

Use the identity: If

p+q+r=0

, then

p^3+q^3+r^3 = 3pqr

.

Explanation:

Let

u = a-b

,

v = b-c

,

w = c-a

. Then

u+v+w = 0

.

Rewrite the sum:

\frac{u^2}{vw} + \frac{v^2}{wu} + \frac{w^2}{uv}

.

Write each term with denominator

uvw

:

\frac{u^3}{uvw} + \frac{v^3}{uvw} + \frac{w^3}{uvw} = \frac{u^3+v^3+w^3}{uvw}

.

Since

u+v+w=0

,

u^3+v^3+w^3 = 3uvw

. So the sum equals

\frac{3uvw}{uvw}=3

.

The original expression is this sum minus 3:

3 - 3 = 0

.

Answer:

A. 0

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