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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

a \neq b \neq c

?

0
3
a + b + c
3 (a−b) (b−c) (c−a)

Solution:

Expression to evaluate:

\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

Step 1: Let

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

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

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

So expression becomes:

x+y+z-3

Step 2: Take LCM of all 3 terms:

\text{LCM}=(a-b)(b-c)(c-a)

Write all terms with common denominator:

x=\frac{(a-b)^2}{(b-c)(c-a)}=\frac{(a-b)^3}{(a-b)(b-c)(c-a)}

y=\frac{(b-c)^2}{(c-a)(a-b)}=\frac{(b-c)^3}{(b-c)(c-a)(a-b)}

z=\frac{(c-a)^2}{(a-b)(b-c)}=\frac{(c-a)^3}{(c-a)(a-b)(b-c)}

Numerators:

\Rightarrow x+y+z=\frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)}

Use identity:

If

x+y+z=0 \Rightarrow x^3+y^3+z^3=3xyz

Here, let:

u=(a-b),\,v=(b-c),\,w=(c-a)

Then:

u+v+w=0

So:

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

Apply:

Numerator

=(a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)

Denominator

=(a-b)(b-c)(c-a)

So:

x+y+z=\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=3

Final Expression

=x+y+z-3=3-3=0

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