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UPSC CDS I 2019 Math Paper

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Question : 46 of 100

Marks: +1, -0

What is

\frac{(x-y)^{3} + (y-z)^{3} + (z-x)^{3}}{3(x-y)(y-z)(z-x)}

equal to?

1
0
$\frac{1}{3}$
3

Solution:

Concept:

Use the algebraic identity: if

a+b+c=0

then

a^3+b^3+c^3 = 3abc

.

Explanation:

Let

a = x-y

,

b = y-z

,

c = z-x

.

Then

a+b+c = (x-y)+(y-z)+(z-x) = 0

.

Thus

a^3+b^3+c^3 = 3abc

.

So

(x-y)^3 + (y-z)^3 + (z-x)^3 = 3(x-y)(y-z)(z-x)

.

Substitute into the expression:

\frac{(x-y)^3 + (y-z)^3 + (z-x)^3}{3(x-y)(y-z)(z-x)} = \frac{3(x-y)(y-z)(z-x)}{3(x-y)(y-z)(z-x)} = 1

.

(Assume denominator is non‑zero; otherwise expression is undefined.)

Answer:

1

(Option A)

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