Find the value of x^{3}-frac{1}{x^{3}} when x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that $x-\frac{1}{x}=a$.We know the algebraic identity $(x-y)^{3}=x^{3}-y^{3}-3xy(x-y)$.Substituting $y = \frac{1}{x}$ into the identity,we g

Vedclass · Accepted Answer

$a^{3}+3a$

Vedclass · Answer

(C) Given that $x-\frac{1}{x}=a$.We know the algebraic identity $(x-y)^{3}=x^{3}-y^{3}-3xy(x-y)$.Substituting $y = \frac{1}{x}$ into the identity,we get:$\left(x-\frac{1}{x}\right)^{3}=x^{3}-\left(\frac{1}{x}\right)^{3}-3(x)\left(\frac{1}{x}\right)\left(x-\frac{1}{x}\right)$.Simplifying the expression:$\left(x-\frac{1}{x}\right)^{3}=x^{3}-\frac{1}{x^{3}}-3\left(x-\frac{1}{x}\right)$.Now,substitute the given value $x-\frac{1}{x}=a$ into the equation:$a^{3}=x^{3}-\frac{1}{x^{3}}-3(a)$.Rearranging the terms to solve for $x^{3}-\frac{1}{x^{3}}$:$x^{3}-\frac{1}{x^{3}}=a^{3}+3a$.

Find the value of $x^{3}-\frac{1}{x^{3}}$ when $x-\frac{1}{x}=a$.

Similar Questions

Number of integral values of $a$ for which both roots of the quadratic equation $x^2 - (2a + 3)x + a^2 + 3a = 0$ lie in the interval $(0, 4)$ is:

$A$ real root of the equation $\log_{4}\{\log_{2}(\sqrt{x+8} - \sqrt{x})\} = 0$ is

Solve the given two equations and choose the correct option.
$I.$ $\frac{3x}{3x+7} - \frac{3x+7}{3x} = 14$
$II.$ $\frac{y}{18y-5} - \frac{18y-5}{y} = 2$

If $a$ and $b$ are odd integers,then the roots of the equation $2ax^2 + (2a + b)x + b = 0, a \ne 0,$ will be

Solve the given equations and select the correct option.
$I.$ $7x + 6y = 110$
$II.$ $4x + 3y = 59$
$III.$ $x + z = 15$

Vedclass Test Series

Exam Paper Generator

Online Exam Module