The rationalising 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 $(x^2 - xy + y^2)$ to obtain $x^3 + y^3 = a + a^{-1}$,which is a rational expression.Substituting the values of $x$ and $y$ into the factor $(x^2 - xy + y^2)$:$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 rationalising factor is $a^{2/3} - 1 + a^{-2/3}$,which is $a^{2/3} + a^{-2/3} - 1$.

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

Similar Questions

$\frac{\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}}}{\sqrt{8+\sqrt{28}}-\sqrt{8-\sqrt{28}}}$ is equal to

$\frac{\sqrt{5/2} + \sqrt{7 - 3\sqrt{5}}}{\sqrt{7/2} + \sqrt{16 - 5\sqrt{7}}} = $

If $(a^m)^n = a^{m^n}$,then the value of $m$ in terms of $n$ is

If $12^{4+2x^2} = (24\sqrt{3})^{3x^2-2}$,then $x$ is equal to

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module