Question

x plus one upon x is equal to 3 square root bulate is x cube + 1 upon x cube​

Answer 1

0

Step-by-step explanation:

Given:

$x + \dfrac{1}{x} = \sqrt{3}$

To calculate:

${x}^{3} + \dfrac{1}{ {x}^{3} }$

What you have to do ?

1. Cubing both side

2. Use Algebra Formula

3. Substitute given value

4. Simplify and answer .... Easy ?

Solution:

$x + \dfrac{1}{x} = \sqrt{3}$

[Cubing both side]

${(x + \frac{1}{x})}^{3} = {( \sqrt{3}) }^{3} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3.x. \frac{1}{x}= 3 \sqrt{3} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x}) = 3 \sqrt{3} \: \: \: [\text{From given}] \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3. \sqrt{3} = 3 \sqrt{3} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 3 \sqrt{3} - 3 \sqrt{3} \\ \\ \boxed{ {x}^{3} + \frac{1}{ {x}^{3} } = 0}$

Therefore, the required answer is 0.

Some Important Algebra Formula

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