Question

if x minus one upon x is equal to 3 then find the value of x cube minus one by x cube equal to

Answer 1

Answer:

It would be equal to 36.

Step-by-step explanation:

Given,

$x-\frac{1}{x}=3$

$\implies (x-\frac{1}{x})^3=27$

Since, (a - b)³ = a³ - 3a²b + 3ab² - b³,

So,

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

$=x^3 - 3x + \frac{3x}{x^2} - \frac{1}{x^3}$

$=x^3 - 3x +\frac{3}{x} - \frac{1}{x^3}$

$=x^3 -\frac{1}{x^3}-3(x-\frac{1}{x})$

$=x^3 - \frac{1}{x^3}-3(3)$

$=x^3 -\frac{1}{x^3}-9$

$\implies x^3 -\frac{1}{x^3}-9 = 27$

$\implies x^3 -\frac{1}{x^3} = 27 + 9 = 36$

Learn more:

https://brainly.in/question/10105256

https://brainly.in/question/10877937

Answer 2

Answer:

$\red { Value \: of \: x^{3}-\left(\frac{1}{x}\right)^{3}} = \green { 36 }$

Step-by-step explanation:

$Given \: x -\frac{1}{x} = 3 \: ---(1)$

$\boxed {\pink {a^{3}-b^{3} = (a-b)^{3} +3ab(a-b)}}$

$x^{3}-\left(\frac{1}{x}\right)^{3}\\ = \left(x-\frac{1}{x}\right)^{3} + 3\times x \times \frac{1}{x}\left(x-\frac{1}{x}\right)$

$=\left(x-\frac{1}{x}\right)^{3} + 3\times \left(x-\frac{1}{x}\right)$

$= 3^{3} +3 \times 3\\= 27 + 9\\=36\: [ From \: (1)]$

Therefore.,

$\red { Value \: of \: x^{3}-\left(\frac{1}{x}\right)^{3}} = \green { 36 }$

•••♪