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UPSC CDS II 2025 Mathematics Paper

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Question : 7 of 100

Marks: +1, -0

If

\frac{1}{a}+\frac{1}{b}=\frac{5}{6}

\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{13}{36}

\frac{1}{a^{3}}+\frac{1}{b^{3}}

equal to?

$31/216$
$35/216$
$37/216$
$41/216$

Solution:

Concept:

Use the algebraic identities for sum of reciprocals:

(x+y)^2 = x^2 + y^2 + 2xy

and

(x+y)^3 = x^3 + y^3 + 3xy(x+y)

. Here

x = \frac{1}{a}

and

y = \frac{1}{b}

.

Explanation:

Let

p = \frac{1}{a} + \frac{1}{b} = \frac{5}{6}

and

q = \frac{1}{a^2} + \frac{1}{b^2} = \frac{13}{36}

.

Square

p

:

p^2 = \frac{25}{36} = q + 2 \cdot \frac{1}{ab}

.

So

\frac{25}{36} = \frac{13}{36} + 2 \cdot \frac{1}{ab}

.

Thus

2 \cdot \frac{1}{ab} = \frac{12}{36} = \frac{1}{3}

, giving

\frac{1}{ab} = \frac{1}{6}

.

Cube

p

:

p^3 = \frac{125}{216} = \left(\frac{1}{a^3} + \frac{1}{b^3}\right) + 3 \cdot p \cdot \frac{1}{ab}

.

Now

3 \cdot p \cdot \frac{1}{ab} = 3 \cdot \frac{5}{6} \cdot \frac{1}{6} = \frac{15}{36} = \frac{5}{12} = \frac{90}{216}

.

Therefore

\frac{1}{a^3} + \frac{1}{b^3} = \frac{125}{216} - \frac{90}{216} = \frac{35}{216}

.

Answer:

\frac{35}{216}

(Option B)

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