UPSC CDS I 2025 Mathematics Paper

Expression to evaluate:
‌

(a−b)2

(b−c)(c−a)

+‌

(b−c)2

(c−a)(a−b)

+‌

(c−a)2

(a−b)(b−c)

−3
Step 1: Let
‌x=‌

(a−b)2

(b−c)(c−a)

‌y=‌

(b−c)2

(c−a)(a−b)

‌z=‌

(c−a)2

(a−b)(b−c)

So expression becomes: x+y+z−3
Step 2: Take LCM of all 3 terms:
LCM=(a−b)(b−c)(c−a)
Write all terms with common denominator:
‌x=‌

(a−b)2

(b−c)(c−a)

=‌

(a−b)3

(a−b)(b−c)(c−a)

‌y=‌

(b−c)2

(c−a)(a−b)

=‌

(b−c)3

(b−c)(c−a)(a−b)

‌z=‌

(c−a)2

(a−b)(b−c)

=‌

(c−a)3

(c−a)(a−b)(b−c)

Numerators:
⇒x+y+z=‌

(a−b)3+(b−c)3+(c−a)3

(a−b)(b−c)(c−a)

Use identity:
If x+y+z=0⇒x3+y3+z3=3xyz
Here, let: u=(a−b),v=(b−c),w=(c−a)
Then: u+v+w=0
So: u3+v3+w3=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=‌

3(a−b)(b−c)(c−a)

(a−b)(b−c)(c−a)

=3
Final Expression =x+y+z−3=3−3=0