If x⁴+1/x⁴ =119, find the value of x³ - 1/x³
Answers
Given : x⁴ + 1/x⁴ = 119
To find : value of x³ - 1/x³
Solution :
We know that
(x² +1/x² )² = (x²)² + (1/x²)² + 2 x² × 1/x²
(x² + 1/x² )² = x⁴ + 1/x⁴ + 2
(x² + 1/x² )² = 119 + 2
(x² + 1/x² )² = 121
(x² + 1/x² )² = 11²
x² + 1/x² = 11……….(1)
[Taking square root of both sides]
Now ,
We know that (x - 1/x)² = x² + 1/x² - 2
(x - 1/x)² = 11 - 2
[From eq 1]
(x - 1/x)² = 9
(x - 1/x)² = 3²
x - 1/x = 3 ………..(2)
On Cubing eq 2 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 = 36
Hence the value of the value of x³ - 1/x³ is 36.
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Given:
x⁴ + 1/x⁴ = 119
To find:
x³ + 1/x³ = ?
Solution:
Identities to be used:
(a + b)² = a² + b² + 2ab
(a - b)² = a² + b² - 2ab
(a - b)³ = a³ - b³ - 3ab(a - b)
(x² + 1/x²)² = x⁴ + 1/x⁴ + 2(x²)(1/x²)
=> (x² + 1/x²)² = 119 + 2
=> (x² + 1/x²)² = 121
=> x² + 1/x² = 11 __(i)
(x - 1/x)² = x² + 1/x² - 2(x)(1/x)
=> (x - 1/x)² = 11 - 2 [from (i)]
=> (x - 1/x)² = 9
=> x - 1/x = 3 __(ii)
(x - 1/x)³ = x³ - 1/x³ - 3(x)(1/x)(x - 1/x)
=> 3³ = x³ - 1/x³ - 3(3) [from (ii)]