Question: If $x²+2 = 2⅔ + 2^{-⅔}$, show that: 2x(x²+3) = 3.


Solution:
Given,

$x² + 2= 2^{\frac{2}{3}} + 2^{-\frac{2}{3}}$

$or, x² = 2^{\frac{2}{3}} - 2 + 2^{-\frac{2}{3}}$

$or, x² = 2^{\frac{2}{3}} -2 + \dfrac{1}{2^{\frac{2}{3}}}$

$or, x² = \left ( 2^{\frac{1}{3}} \right )^2 - \left ( 2 * 2^{\frac{1}{3}} * \dfrac{1}{2^{\frac{1}{3}}} \right ) + \left ( \dfrac{1}{2^{\frac{1}{3}}} \right )^2$

$or, x² = \left ( 2^{\frac{1}{3}} - \dfrac{1}{2^{\frac{1}{3}}} \right )^2$

$or, x = \left ( 2^{\frac{1}{3}} - \dfrac{1}{2^{\frac{1}{3}}} \right ) $ - (i)

[ Cubing both sides of equation (i) ]

$or, x³ = \left ( 2^{\frac{1}{3}} - \dfrac{1}{2^{\frac{1}{3}}} \right )^3 $

[ (a-b)³ = a³-b³ -3ab(a-b) ]

$or, x³ =  \left ( 2^{\frac{1}{3}} \right )^3  -  \left ( \dfrac{1}{2^{\frac{1}{3}}} \right )^3 - 3*2^{\frac{1}{3}} * \dfrac{1}{2^{\frac{1}{3}}} \left (2^{\frac{1}{3}} - \dfrac{1}{2^{\frac{1}{3}}} \right )$

[ From (i)  $\left (2^{\frac{1}{3}} - \dfrac{1}{2^{\frac{1}{3}}} \right ) = x$ ]

$or, x³ = 2 - \dfrac{1}{2} - 3*1*(x)$

$or, x³ = 2 - \dfrac{1}{2} - 3x$

$or, x³ = \dfrac{4 - 1 -6x}{2}$

$or, 2x³ = 3 - 6x$

$or, 2x³ +6x = 3$

$\therefore, 2x(x²+3) = 3$
#proved

Related Notes and Solutions:

Here is the website link to the notes of Indices.

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