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UPSC CDS 2 2019 Math Paper

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Question : 98 of 100

Marks: +1, -0

What is the value of

\frac{{{{\left( {x - y} \right)}^3}\; + \;{{\left( {y - z} \right)}^3}\; + \;{{\left( {z - x} \right)}^3}}}{{9\left( {x - y} \right)\left( {y - z} \right)\left( {z - x\;} \right)}}

?

0
$1/3$
$1/9$
1

Solution:

Concept:

Use the 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)

.

Check their sum:

(x - y) + (y - z) + (z - x) = 0

.

Thus,

a + b + c = 0

.

Applying the identity,

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

.

Substitute into the given expression:

\frac{(x - y)^{3} + (y - z)^{3} + (z - x)^{3}}{9(x - y)(y - z)(z - x)} = \frac{3(x - y)(y - z)(z - x)}{9(x - y)(y - z)(z - x)} = \frac{3}{9} = \frac{1}{3}

.

Answer:

\frac{1}{3}

(Option B).

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