If x-sqrt{3}-sqrt{2}=0 and y-sqrt{3}+sqrt{2}= | vedclass

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

Question

(B) Given equations are $x = \sqrt{3} + \sqrt{2}$ and $y = \sqrt{3} - \sqrt{2}$.First,calculate $(x - y)$:$x - y = (\sqrt{3} + \sqrt{2}) - (\sqrt{3} -

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Given equations are $x = \sqrt{3} + \sqrt{2}$ and $y = \sqrt{3} - \sqrt{2}$.First,calculate $(x - y)$:$x - y = (\sqrt{3} + \sqrt{2}) - (\sqrt{3} - \sqrt{2}) = 2\sqrt{2}$.Next,calculate $(xy)$:$xy = (\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$.We need to evaluate $(x^3 - 20\sqrt{2}) - (y^3 + 2\sqrt{2}) = x^3 - y^3 - 22\sqrt{2}$.Using the identity $x^3 - y^3 = (x - y)^3 + 3xy(x - y)$:$x^3 - y^3 = (2\sqrt{2})^3 + 3(1)(2\sqrt{2})$$x^3 - y^3 = 16\sqrt{2} + 6\sqrt{2} = 22\sqrt{2}$.Substituting this into the expression:$22\sqrt{2} - 22\sqrt{2} = 0$.

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:

Similar Questions

Find the value of $(a+b+c)^{3}-3(b+c)(c+a)(a+b)$ if $a=5, b=3$ and $c=2$.

The roots of the equation $\sqrt{3x + 1} + 1 = \sqrt{x}$ are

If the roots of the equation $ax^2 + bx + c = 0$ are $\alpha$ and $\beta$,then the value of $\alpha\beta^2 + \alpha^2\beta + \alpha\beta$ will be:

Solve the given two equations and select the correct answer from the given options.
$I.$ $\quad x = \sqrt[3]{2197}$
$II.$ $\quad y^2 = 169$

For what value of $k$ can the quadratic polynomial $3z^{2} + 5z + k$ be factored into a product of real linear factors?

Vedclass Test Series

Exam Paper Generator

Online Exam Module