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UPSC CDS 1 2022 Math Paper

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Question : 2 of 100

Marks: +1, -0

If

a+b=2,\;\frac{1}{a}+\;\frac{1}{b}=2

then what is the value of

a^{3}+b^{3}

?

2
4
6
8

Solution: 👈: Video Solution

Given:

a+b=2,\;\frac{1}{a}+\;\frac{1}{b}=2

Formual Used:

a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-a b)

Calculation:

We have

a+b=2 \;\; \cdots-(i)

And

\;\frac{1}{a}+\;\frac{1}{b}=2

\Rightarrow a + b = 2 ab

\Rightarrow 2=2 ab \;\; [\;\text{ from (i) }\;]

\Rightarrow ab = 1 \;\; \cdots \;\text{ (ii) }\;

Squaring the equation (i), we get

a^{2}+ b^{2}+2 ab=4

\Rightarrow a^{2}+ b^{2}+2 \times 1=4

\Rightarrow a^{2}+ b^{2}=2

According to the formula used

a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-a b)

And from equation (i), (ii) and (iii), we get

\Rightarrow a^{3}+ b^{3}=2(2-1)

\Rightarrow a^{3}+ b^{3}=2

\therefore

The required value of

a^{3}+b^{3}

is 2 .

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