If $x-\sqrt{3}-\sqrt{2}=0$ and $y-\sqrt{3}+\sqrt{2}=0,$ then the value of $(x^{3}-20\sqrt{2})-(y^{3}+2\sqrt{2})$ is:

  • A
    $1$
  • B
    $0$
  • C
    $2$
  • D
    $3$

Explore More

Similar Questions

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + 2x - 5 = 0$ and the equation $x^3 + bx^2 + cx + d = 0$ has roots $2\alpha + 1, 2\beta + 1, 2\gamma + 1$,then the value of $|b + c + d|$ is (where $b, c, d$ are constants):

Difficult
View Solution

If $\alpha$ and $\beta$ are the roots of the equation $7x^{2}-3x-2=0$,then the value of $\frac{\alpha}{1-\alpha^{2}}+\frac{\beta}{1-\beta^{2}}$ is equal to

If $\alpha$ and $\beta$ are roots of the equation $x^2 - 4\sqrt{2}kx + 2e^{4\ln k} - 1 = 0$ for some $k$, and $\alpha^2 + \beta^2 = 66$, then $\alpha^3 + \beta^3$ is equal to: (in $\sqrt{2}$)

Difficult
View Solution

Solve the given two equations and select the correct option.
$I.$ $x = \frac{\sqrt{256}}{\sqrt{576}}$
$II.$ $3y^2 + y - 2 = 0$

If $\alpha, \beta$ are the roots of the equation $x^2 - 2x + 4 = 0$,then the value of $\alpha^n + \beta^n$ is

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo