Math, asked by kshitijtayade9461, 10 months ago

If x + 1/x = 3, Calculate x² + 1/x², x³ + 1/x³, and x⁴+1/x⁴

Answers

Answered by nikitasingh79
9

Given : x + 1/x = 3

To find : value of x² + 1/x², x³ + 1/x³, and x⁴+1/x⁴

 

Solution :  

We have  x + 1/x = 3 ……..(1)

On squaring eq 1 both sides,  

(x + 1/x)²   = 3²

By Using Identity : (a + b)²  = a² + b² + 2ab

x² + 1/x² + 2 x × 1/x = 9

x² + 1/x² + 2 = 9

x² + 1/x² =  9 - 2

x² + 1/x² = 7 ………….(2)

 

On squaring eq 2 both sides,  

(x² +1/x² )² = 7²

(x²)² + (1/x²)² + 2 x² × 1/x² =7²

x⁴ + 1/x⁴ + 2 = 49

x⁴ + 1/x⁴ = 49 - 2

x⁴ + 1/x⁴ = 47

On Cubing eq 1 both sides :

(x + 1/x)³  = 3³  

By Using Identity : (a + b)³  = a³ + b³ + 3ab(a + b)

(x)³ + (1/x)³ + 3 × x× 1/x (x + 1/x) = 27

 x³ + 1/x³ + 3(x + 1/x) = 27

x³ + 1/x³ + 3(3) = 27

x³ + 1/x³ + 9 =  27

x³ + 1/x³ =  27 - 9

x³ + 1/x³ =  18

Hence the value of the value of x² + 1/x² is 7 , x³ + 1/x³ is   18 &   x⁴ + 1/x⁴ is 47.

HOPE THIS ANSWER WILL HELP YOU…..

 

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Answered by Anonymous
6

Given:

x + 1/x = 3

To find:

x³ + 1/x³ = ?

x² + 1/x² = ?

x⁴ + 1/x⁴ = ?

Solution:

Identities to be used:

(a + b)² = a² + b² + 2ab

(a + b)³ = a³ + b³ + 3ab(a + b)

(x + 1/x)² = x² + 1/x² + 2(x)(1/x)

=> 3² = x² + 1/x² + 2

=> x² + 1/x² = 7

=> x² + 1/x² = 7 __(i)

(x² + 1/x²)² = x⁴ + 1/x⁴ + 2(x²)(1/x²)

=> x⁴ + 1/x⁴ = 7² - 2 [from (i)]

=> x⁴ + 1/x⁴ = 49 - 2

=> x⁴ + 1/x⁴ = 47 __(ii)

(x + 1/x)³ = x³ + 1/x³ + 3(x)(1/x)(x + 1/x)

=> 3³ = x³ + 1/x³ + 3(3) [from (ii)]

=> x³ + 1/x³ = 18

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