If x + 1/x = 3, Calculate x² + 1/x², x³ + 1/x³, and x⁴+1/x⁴
Answers
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.
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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)]