Question

if x+l/x=3 then x3+1/x3​

Answer 1

Answer :

36

Step by step explaination:

: X - 1/x = 3

: X - 1/x = 3 On cubing both sides ; (x - 1/x) = (3)³

: X - 1/x = 3 On cubing both sides ; (x - 1/x) = (3)³ Using Identity : (a -b) = a° - b³ - 3 ab ( a - b)

: X - 1/x = 3 On cubing both sides ; (x - 1/x) = (3)³ Using Identity : (a -b) = a° - b³ - 3 ab ( a - b) → x³ - 1/x° 3 (x - 1/ x ) = 27

: X - 1/x = 3 On cubing both sides ; (x - 1/x) = (3)³ Using Identity : (a -b) = a° - b³ - 3 ab ( a - b) → x³ - 1/x° 3 (x - 1/ x ) = 27 - x° - 1/x³ - 3 (3) = 27 - x° - 1/x - 9 = 27

: X - 1/x = 3 On cubing both sides ; (x - 1/x) = (3)³ Using Identity : (a -b) = a° - b³ - 3 ab ( a - b) → x³ - 1/x° 3 (x - 1/ x ) = 27 - x° - 1/x³ - 3 (3) = 27 - x° - 1/x - 9 = 27 → x³ - 1/ x = 27 + 9 - x° - 1/x = 36

Answer 2

Given :-

$x + \dfrac{1}{x} = 3$

To find :-

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

Identity to use :-

(a+b)³ = a³ + 3ab(a+b) + b³

(or)

a³+b³ = (a+b)(a²-ab+b²)

Using identity (a+b)³ identity,

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

Substituting the value of x+1/x as 3, and cancelling x and 1/x as it's multiplication,

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

$27 = {x}^{3} + 9 + \dfrac{1}{ {x}^{3} }$

$27 - 9 = {x}^{3} + \dfrac{1}{ {x}^{3} }$

$18 = {x}^{3} + \dfrac{1}{ {x}^{3} }$

Using identity a³+b³

${x}^{3} + \dfrac{1}{ {x}^{3} } = \bigg(x + \dfrac{1}{x} \bigg) \bigg( {x}^{2} - (x)( \frac{1}{x} ) + \dfrac{1}{ {x}^{2} } \bigg)$

Substituting the value of x+1/x as 3 and cancelling (-x) and (1/x),

${x}^{3} + \dfrac{1}{ {x}^{3} } = (3) \bigg( {x}^{2} - 1 + \dfrac{1}{ {x}^{2} } \bigg)$

We have to find the value of x² and 1/x²

By using the identity (a+b)² = a² + 2ab + b² let us find the value,

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

Cancelling x and 1/x and substituting the value of x+1/x as 3,

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

$9 = {x}^{2} + \dfrac{1}{ {x}^{2} } + 2$

$9 - 2 = {x}^{2} + \dfrac{1}{ {x}^{2} }$

$7 = {x}^{2} + \dfrac{1}{ {x}^{2} }$

Now let us substitute the value of x²+1/x²

${x}^{3} + \dfrac{1}{ {x}^{3} } = 3(7 - 1)$

${x}^{3} + \dfrac{1}{ {x}^{3} } = 3(6)$

${x}^{3} + \dfrac{1}{ {x}^{3} } = 18$

Hence,

${x}^{3} + \dfrac{1}{ {x}^{3} } = 18$