Question

If a =1-√2,Find the value of a-1/a^3

Answer 1

a=1-√2
1/a=1/(1-√2)
or, 1/a=(1+√2)/(1-√2)(1+√2)
or, 1/a=(1+√2)/(1-2)
or, 1/a=-(1+√2)
∴, (a-1/a)³=[(1-√2)-{-(1+√2)}]³=(1-√2+1+√2)³=2³=8

Answer 2

Answer:

The value of $a-\frac {1}{a^3}$ is 8.

To find:  

The value of $a-\frac {1}{a^3}$

Solution:

Given:

$\begin{array} { l } { \mathrm { a } = 1 - \sqrt { 2 } } \\\\ { \frac { 1 } { a } = \frac { 1 } { 1 - \sqrt { 2 } } } \end{array}$

$\begin{array} { c } { = \frac { 1 } { 1 - \sqrt { 2 } } \times \frac { ( 1 + \sqrt { 2 } ) } { 1 + \sqrt { 2 } } } \\\\ { = \frac { ( 1 + \sqrt { 2 } ) } { 1 ^ { 2 } - \sqrt { 2 } ^ { 2 } } } \end{array}$

$\begin{aligned} & = \frac { ( 1 + \sqrt { 2 } ) } { 1 - 2 } \\\\ & = \frac { ( 1 + \sqrt { 2 } ) } { - 1 } \\\\ & = - ( 1 + \sqrt { 2 } ) \end{aligned}$

$\begin{array} { l } { \frac { 1 } { a } = - 1 - \sqrt { 2 } } \\\\ { a - \frac { 1 } { a ^ { 3 } } } \end{array}$

$\begin{aligned} = & ( 1 - \sqrt { 2 } - ( - 1 - \sqrt { 2 } ) ) ^ { 3 } \\\\ & = ( 1 - \sqrt { 2 } + 1 + \sqrt { 2 } ) ^ { 3 } \end{aligned}$

$=2^3$

$= 8$

Therefore,

$\mathrm { a } - \frac { 1 } { \mathrm { a } ^ { 3 } } = 8$

Thus, the value of $a - \frac { 1 } { a ^ { 3 } }$ is 8.