Question

Please answer me fast, very fast ​

Answer 1

$\sf{ \tt \underline{ \pink{Ea} \blue{sy} \: \: \purple{Q}uestio \red{n}}} :)$

Given:

$a + \dfrac{1}{a} = \sqrt{3}$

To Prove:

${a}^{3} + \dfrac{1}{ {a}^{3} } = 0$

How to prove ?

1. As you see cube in question, It's mean we will cube both side.

2. As we are cubing, Formula will be use.

3. Where is Formula ?

$\tt \: Here \: it's \: \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)]$

4. Just Put the value in Formula.

4. Sound Difficult, No way !

Solution:

L.H.S

$a + \dfrac{1}{a} = \sqrt{3}$

[Cubing both side]

$\implies {(a + \frac{1}{a}) }^{3} = {( \sqrt{3})}^{3} \\ \\ \implies {a}^{3} + \frac{1}{ {a}^{3} } + 3.a. \frac{1}{a} (a + \frac{1}{a}) = 3 \sqrt{3} \\ \\ \implies {a}^{3} + \frac{1}{ {a}^{3} } + 3. \cancel{a}. \frac{1}{\cancel{a}} ( \sqrt{3} ) = 3 \sqrt{3} \\ \\ \implies {a}^{3} + \frac{1}{ {a}^{3} } + 3 \sqrt{3} = 3 \sqrt{3} \\ \\ \implies {a}^{3} + \frac{1}{ {a}^{3} } = 3 \sqrt{3} - 3 \sqrt{3} \\ \\ \implies \boxed{{a}^{3} + \frac{1}{ {a}^{3} } = 0} \\ \\ \therefore \tt \: L.H.S = \: R.H.S \: \:\: [Proved]$