Question

if (a+1/a)²=3 and a is not equal to 0 then show that a³+1/a³=0​

Answer 1

${\underline{{\text{Given:}}}}$

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

${\underline{{\text{To Prove:}}}}$

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

${\underline{{\text{Solution:}}}}$

First we should find the value of $a + \dfrac{1}{a}$

So,

${(a + \frac{1}{a}) }^{2} = 3 \\ \\ a + \frac{1}{a} = \pm \sqrt{3} ......(1) \:\:\: \:\:[\text{Using+value]} \\ \\ [\text{Cubing both side}] \\ \\ {(a + \frac{1}{a}) }^{3} = {( \sqrt{3} )}^{3} \\ \\ {a}^{3} + \frac{1}{ {a}^{3} } + 3.a. \frac{1}{a} (a + \frac{1}{a}) = 3 \sqrt{3} \\ \\ {a}^{3} + \frac{1}{ {a}^{3} } + 3.\cancel{a}. \frac{1}{ \cancel{a}} ( \sqrt{3} ) = 3 \sqrt{3} \: \: \: [\text{From ...1}] \\ \\ {a}^{3} + \frac{1}{ {a}^{3} } + 3 \sqrt{3} = 3 \sqrt{3} \\ \\ {a}^{3} + \frac{1}{ {a}^{3} } = 3 \sqrt{3} - 3 \sqrt{3} \\ \\ {a}^{3} + \frac{1}{ {a}^{3} } = 0 \\ \\ \therefore \: \: {a}^{3} + \frac{1}{ {a}^{3} } = 0 \: \: \: \: \text{[Proved]}$

${\underline{{\text{Important Algebra Formula}}}}$

${(x + y)}^{2}={x}^{2}+{y}^{2}+2xy\\ \\{(x - y)}^{2}={x}^{2}+{y}^{2}-2xy\\ \\{(x+y)}^{2}= (x - y) {}^{2}+4xy\\ \\{(x-y)}^{2}=(x+y){}^{2}-4xy\\ \\ (x + y)^{2}+(x-y)^{2}=2( {x}^{2}+{y}^{2} )\\ \\(x+y)^{2}- (x-y) {}^{2}=4xy\\ \\ {(x + y)}^{3}={x}^{3}+{y}^{3}+ 3xy(x + y) \\ \\(x - y)^{3}={x}^{3}-{y}^{3}- 3xy(x - y)$

Answer 2

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