Question

If X2+1/x2=23, then find the Value of X+1/x and X3+1/X3​

Answer 1

Answer:

x + 1/x = 5

x³ + 1/x³ = 110

Step-by-step explanation:

Given : ${x}^{2} + \dfrac{1}{ {x}^{2} } = 23$

First, we have to find $x + \dfrac{1}{x} .$

Squaring of $x + \dfrac{1}{x}$, we get

→ ${(x + \dfrac{1}{x} )}^{2} = {x}^{2} + \dfrac{1}{ {x}^{2} } + 2.x. \dfrac{1}{x}$

[Identity used : (a + b)² = a² + b² + 2ab]

Putting known values, we get

→ ${(x + \dfrac{1}{x} )}^{2} = 23 + 2$

→ ${(x + \dfrac{1}{x}) }^{2} = 25$

→ $x + \dfrac{1}{x} = \sqrt{25}$

→ $x + \dfrac{1}{x} =$ 5

Now,

Cubing of $x + \dfrac{1}{x}$, we get

→ ${(x + \dfrac{1}{x} )}^{3} = {x}^{3} + \dfrac{1}{ {x}^{3} } + 3.x. \dfrac{1}{x} (x + \dfrac{1}{x} )$

[Identity used : (a + b)³ = a³ + b³ + 3ab(a + b)]

Putting known values, we get

→ ${(5)}^{3} = {x}^{3} + \dfrac{1}{ {x}^{3} } + 3(5)$

→ $125 = {x}^{3} + \dfrac{1}{ {x}^{3} } + 15$

→ ${x}^{3} + \dfrac{1}{ {x}^{3} } = 125 - 15$

→ ${x}^{3} + \dfrac{1}{ {x}^{3} } =$ 110