Question

Simplify $\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}.$

Answer 1

Given :    $\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}.$    or 2∛9 / ( 1 + ∛3  + ∛9)

To find : Simplify  ( rationalize)

Solution:

$\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}.$

2∛9 / ( 1 + ∛3  + ∛9)

a = ∛3   ,    b =  1

=> a² =  ∛9    ab = ∛3   b² = 1

as we know that (a - b) (a² + ab + b²)   = a³ - b³

Hence multiplying numerator & denominator with a - b  i.e.  ∛3  - 1

= 2∛9  (   ∛3 - 1 ) / ( 3 - 1 )

= 2∛9  (   ∛3 - 1 )/ 2

= ∛9  (   ∛3 - 1 )

= ∛27 - ∛9

= 3 - ∛9

2∛9 / ( 1 + ∛3  + ∛9)  = 3 - ∛9

$\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9} = 3 - \sqrt[3]9$

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