Question

if x+1/x =2, then x³+1/x³=​

Answer 1

Step-by-step explanation:

Let p (x) = x + 1 / x = 2

= x + 1 = 2x

= 2x - x = 1

or, x = 1.

let g(x) = x^3 + 1/x^3

since, x = 1

therefore,

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

= 1 + 1 / 1

= 2 / 1

= 2

So, the answer to your question is 2.

You can also make it this way -

x + 1/ x = 2

Cubing both sides, we get -

x^3 + 1/x^3 + 3×x×1/x (x + 1/x) = 8

= x^3 + 1/x^3 + 3 (x + 1/x) = 8

= x^3 + 1/x^3 + 3 × 2 = 8 (since, x + 1/ x = 2)

= x^3 + 1/x^3 = 8 - 6

or, x^3 + 1/x^3 = 2.

Hope this answer helps you out.

Answer 2

★ᏀᏆᏙᎬΝ:-

$\tt x + \dfrac{1}{x} = 2$

★Ꭲᝪ ᖴᏆᑎᗞ:-

$\tt x^3 + \dfrac{1}{x^3}$

★ꌗꂦ꒒ꀎ꓄ꀤꂦꈤ:-

We know that,

$\boxed{ \tt (a+b)^3 = a^3 +b^3 + 3ab(a+b)}$

Using this identity,

$\tt \Bigg ( x + \dfrac{1}{x} \Bigg )^3 =(x)^3 + \Bigg ( \dfrac{1}{x} \Bigg )^3 + 3 \times x \times \dfrac{1}{x} \Bigg ( x + \dfrac{1}{x} \Bigg )$

$\tt \implies (2)^3 =x^3 + \dfrac{1}{x^3} + 3(2)$

$\tt \implies 8 =x^3 + \dfrac{1}{x^3} + 6$

$\tt \implies 8-6 =x^3 + \dfrac{1}{x^3}$

$\tt \implies 2=x^3 + \dfrac{1}{x^3}$

$\boxed{\tt x^3 + \dfrac{1}{x^3}=2}$