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UPSC CDS 1 2023 Math Paper

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Question : 39 of 100

Marks: +1, -0

If

x = \frac{\sqrt{3}+1}{\sqrt{3}-1}

y = \frac{\sqrt{3}-1}{\sqrt{3}+1}

then what is the value

x^{3} - y^{3}

?

60
45√3
30√3
90

Solution:

Concept:

Use identity:

a^{3} - b^{3} = (a - b)^{3} + 3ab(a - b)

.

Also simplify

x

and

y

to find

x - y

and

xy

.

Explanation:

Given

x = \frac{\sqrt{3}+1}{\sqrt{3}-1}

and

y = \frac{\sqrt{3}-1}{\sqrt{3}+1}

.

First, compute

xy

:

xy = \frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}-1}{\sqrt{3}+1} = 1

.

Thus,

xy = 1

.

Now compute

x - y

:

x - y = \frac{\sqrt{3}+1}{\sqrt{3}-1} - \frac{\sqrt{3}-1}{\sqrt{3}+1}

.

Take common denominator

(\sqrt{3}-1)(\sqrt{3}+1) = (\sqrt{3})^2 - 1^2 = 3 - 1 = 2

:

x - y = \frac{(\sqrt{3}+1)^2 - (\sqrt{3}-1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)}

.

Calculate squares:

(\sqrt{3}+1)^2 = 3 + 1 + 2\sqrt{3} = 4 + 2\sqrt{3}

, and

(\sqrt{3}-1)^2 = 3 + 1 - 2\sqrt{3} = 4 - 2\sqrt{3}

.

So numerator =

(4 + 2\sqrt{3}) - (4 - 2\sqrt{3}) = 4\sqrt{3}

.

Thus

x - y = \frac{4\sqrt{3}}{2} = 2\sqrt{3}

.

Now apply identity for

x^{3} - y^{3}

:

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

.

Substitute

x - y = 2\sqrt{3}

and

xy = 1

:

x^{3} - y^{3} = (2\sqrt{3})^{3} + 3(1)(2\sqrt{3}) = 8 \times 3\sqrt{3} + 6\sqrt{3} = 24\sqrt{3} + 6\sqrt{3} = 30\sqrt{3}

.

Answer:

30\sqrt{3}

, which corresponds to option C.

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