Question

(a+1/a) ²=3 ,a³+1/a³=0 prove this​

Answer 1

Answer:

a³ + 1/a³ = 0

Step-by-step explanation:

⇒ (a + 1/a)² = 3

⇒ (a + 1/a) = √3

       Cube on both sides:

⇒ (a + 1/a)³ = (√3)³

⇒ a³ + (1/a)³ + 3(a)(1/a)(a + 1/a) = 3√3

⇒ a³ + 1/a³ + 3(1))(a + 1/a) = 3√3

⇒ a³ + 1/a³ + 3(√3) = 3√3    [from above]

⇒ a³ + 1/a³ = 3√3 - 3√3

⇒ a³ + 1/a³ = 0         proved

Answer 2

Given:-

• $\sf{\bigg(a + \dfrac{1}{a}\bigg)^{2}} = 3\ \ and \ \ a\neq 0$.

To Prove:-

• $\sf{a^{3} + \dfrac{1}{a^{3}}} = 0$

Proof:-

Since, It is given that  $\sf{\bigg(a + \dfrac{1}{a}\bigg)^{2}} = 3\ \ and \ \ a\neq 0$ .

So, Solving it further we get,

⇒ $\rm{\bigg(a + \dfrac{1}{a}\bigg)^{2}} = 3$

⇒ $\rm{\sqrt{\bigg(a + \dfrac{1}{a}\bigg)^{2}}} = \sqrt{3}$                [ Squaring both sides ]

⇒ $\rm{a+ \dfrac{1}{a}} = \sqrt{3}$

Now, Cubing both sides we get,

⇒ $\rm{\bigg(a+ \dfrac{1}{a}\bigg)^{3}} = (\sqrt{3})^{3}$

⇒ $\rm{a^{3}+ \dfrac{1}{a^{3}} + 3 \times a \times \dfrac{1}{a}(a + \dfrac{1}{a})} = 3\sqrt{3}$

⇒ $\rm{a^{3}+ \dfrac{1}{a^{3}} + 3(\sqrt{3})} = 3\sqrt{3}$                [ Since, $\rm{a+ \dfrac{1}{a}} = \sqrt{3}$ ]

⇒ $\rm{a^{3} + \dfrac{1}{a^{3}}} = 3\sqrt{3} - 3\sqrt{3}$

⇒ $\tt{a^{3} + \dfrac{1}{a^{3}}} = 0$

                             -------------- Hence Proved

Some Important terms:-

• $\sf{(a + b)^{2} = a^{2} + b^{2} + 2ab}$

• $\sf{(a - b)^{2} = a^{2} + b^{2} - 2ab}$

• $\sf{(a - b)^{3} = a^{3} - b^{3} - 3ab(a - b)}$