Question

If a cube-3a +1 =0 find a+1/a (it's the question please help)

Answer 1

Answer:

Here,

a³-3a+1=0

=>a³+1=3a

=>(a+1)(a²-a+1)=3a

=>a+1=3a/(a²-a+1)

=>a+1/a=3/(a²-a+1)

Therefore,a+1/a=3/(a²-a+1)

Answer 2

Using $(a + b)^3 = (a^3 + 3a b(a + b) + b^3)$

Step-by-step explanation:

Given, $(3a + 1)^3 = 0$

$a^3 + 1 = 3a$

⇒ $(a + 1) \times (a^2 - a + 1) = 3a$

⇒ $a + 1 = \frac {3a}{a^2 - a + 1}$

⇒ $a + \frac {1}{a} = \frac {3}{a^2 - a + 1}$

Hence, $a + \frac {1}{a} = \frac {3}{a^2 - a + 1}$ is Answer.