Question

if
$x + \frac{1}{x} = 2$
,then
${x}^{3} + \frac{1}{{x}^{3} }$
​

Answer 1

Answer:

$\large\bold\red{2}$

Step-by-step explanation:

Given,

$x + \frac{1}{x} = 2 \: \: \: \: \: .......(1)$

Squaring both the sides,

We get,

$= > {(x + \frac{1}{x} ) }^{2} = {2}^{2} \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times x \times \frac{1}{x} = 4 \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 4 \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = 2 \: \: \: \: \: .......(2)$

Now,

Cubing both the sides in eqn (1),

We get,

$= > {(x + \frac{1}{x} )}^{3} = {2}^{3} \\ \\ = > {x}^{3} + {( \frac{1}{x} )}^{3} + 3 \times x \times \frac{1}{x} (x + \frac{1}{x} ) = 8 \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x} ) = 8 \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 2 = 8 \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } + 6 = 8 \\ \\ = > \bold{ {x}^{3} + \frac{1}{ {x}^{3} } = 2 }$

Answer 2

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