Question

If x=3+root8findthevalueofx^3-1/x^3

Answer 1

Given numeric value of x is 3 + √8.

$\implies x = 3+\sqrt8 \\ \\ \implies \dfrac{1}{x} = \dfrac{1}{3 + \sqrt{8} }$

Using Rationalisation : Multiply the numerator & denominator by current denominator with opposite sign( between 3 & √8 ).

It means we have to multiply by 3 - √8 .

Now,

$\implies \dfrac{1}{x} = \dfrac{1}{3 + \sqrt{8} } \times \dfrac{3 - \sqrt{8} }{3 - \sqrt{8} } \\ \\ \implies \dfrac{1}{x} = \dfrac{3 - \sqrt{8} }{(3 + \sqrt{8})(3 - \sqrt{8} ) }$

From the properties of expansion : -

• ( a + b )( a - b ) = a^2 - b^2

Here, the above denominator can be solved by using this property.

According to the property, ( 3 + √8 )( 3 - √8 ) = ( 3 )^2 - ( √8 )^2 = 9 - 8 = 1

Thus,

$\implies \dfrac{1}{x} = \dfrac{3 - \sqrt{8} }{1 } \\ \\ \implies \dfrac{1}{x} = 3 - \sqrt{8}$

Hence,

• x = 3 + √8
• 1 / x = 3 - √8

On cubing both sides of x : -

= > x^3 = ( 3 + √8 )^3

= > x^3 = ( 3 )^3 + ( √8 )^3 + 3( 3 + √8 )( 3√8 ) { formula : ( a + b )^3 = a^3 + b^3 + 3ab( a + b ) }

= > x^3 = 27 + 8√8 + 27√8 + 72

= > x^3 = 99 + 35√8

On cubing both sides of 1 / x : -

= > ( 1 / x )^3 = ( 3 - √8 )^3

= > 1 / x^3 = ( 3 )^3 - ( √8 )^3 - 3( 3√8 )( 3 - √8 )

= > 1 / x^3 = 27 - 8√8 - 27√8 + 72

= > 1 / x^3 = 99 - 35√8

Thus,

= > x^3 - 1 / x^3 = 99 + 35√8 - ( 99 - 35√8 )

= > x^3 - 1 / x^3 = 99 + 35√8 - 99 + 35√8 = 70√8 = 70√( 2^2 x 2 ) = ( 70 x 2 )√2 = 140√2

Hence the required numeric value of x^3 - 1 / x^3 is 140√2.