Question

If a+1/a=root 3 then what is sum of a cube and its resiprocal

Answer 1

The answer is given below :

Given,

$a + \frac{1}{a} = \sqrt{3} \\ \\ now \: taking \: cube \: \: we \: \: get \\ \\ ({a}^{3} + \frac{1}{ {a}^{3} } ) + 3 \times a \times \frac{1}{a} \times (a + \frac{1}{a} ) = { (\sqrt{3}) }^{3} \\ \\ or \: \:( {a}^{3} + \frac{1}{ {a}^{3} } ) + 3 \sqrt{3} = 3 \sqrt{3} \\ \\ cancelling \: \: 3 \sqrt{3} \: \: we \: \: get \\ \\ {a}^{3} + \frac{1}{ {a}^{3} } = 0$

Thank you for your question.

Answer 2

Hey!

Here is your answer >>>

a + 1/a = root 3!.

We have to find a^3+(1/a)^3 =

Let us whole cube the equation on both sides!.

(a + 1/a)^3 = a^3 + (1/a)^3 +3a(1/a)(a+1/a)

(root3)^3 = a^3 + (1/a)^3 + 3(root3)

$3 \sqrt{3} = {a}^{3} + { \frac{1}{a} }^{3} + 3( \sqrt{3} )$

a^3 + (1/a)^3 =
$3 \sqrt{3} - 3 \sqrt{3}$

= 0