The rationalizing factor of a^{1/3} + a^{-1/3 | vedclass

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

Question

(D) Let $x = a^{1/3}$ and $y = a^{-1/3}$.We know the algebraic identity $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$.Here,$x^3 = (a^{1/3})^3 = a$ and $y^3 =

Vedclass · Accepted Answer

$a^{2/3} + a^{-2/3} - 1$

Vedclass · Answer

(D) Let $x = a^{1/3}$ and $y = a^{-1/3}$.We know the algebraic identity $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$.Here,$x^3 = (a^{1/3})^3 = a$ and $y^3 = (a^{-1/3})^3 = a^{-1}$.To rationalize the expression $(x + y)$,we need to multiply it by $(x^2 - xy + y^2)$.Substituting the values of $x$ and $y$:$x^2 = (a^{1/3})^2 = a^{2/3}$$y^2 = (a^{-1/3})^2 = a^{-2/3}$$xy = a^{1/3} 	imes a^{-1/3} = a^{1/3 - 1/3} = a^0 = 1$.Therefore,the rationalizing factor is $a^{2/3} + a^{-2/3} - 1$.

The rationalizing factor of $a^{1/3} + a^{-1/3}$ is:

Similar Questions

$\sqrt{2+\sqrt{5}-\sqrt{6-3 \sqrt{5}+\sqrt{14-6 \sqrt{5}}}}$ is equal to

$\frac{1}{\sqrt{11 - 2\sqrt{30}}} - \frac{3}{\sqrt{7 - 2\sqrt{10}}} - \frac{4}{\sqrt{8 + 4\sqrt{3}}} = $

Evaluate: $\sqrt{3 + \sqrt{5}} - \sqrt{2 + \sqrt{3}}$

$\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}} - \frac{1}{\sqrt{5 + 2\sqrt{6}}} = $

If $a^{1/x} = b^{1/y} = c^{1/z}$ and $b^2 = ac$,then $x + z = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module