Q- if x square + 1/ x square = 83 then find the value of x cube - 1/ x cube

x2 + 1/x2 = 83

x2 + 1/x2 - 2 = 83 - 2

(x - 1/x)2 = 81

(x - 1/x)2 = 92

x - 1/x = 9

Now,

(x - 1/x)3 = 93

x3 - 1/x3 - 3*x*1/x (x - 1/x) = 729

x3 - 1/x3 - 3*x*1/x (x - 1/x) = 729

now substituting the value of x - 1/x

x3 - 1/x3 - 3*9 = 729

x3 - 1/x3 - 27 = 729

x3 - 1/x3 = 729 + 27

x3 - 1/x3 = 756

• 32

x2 + 1/x2 = 83

x2 + 1/x2 - 2 = 83 - 2

(x - 1/x)2 = 81

(x - 1/x)2 = 92

x - 1/x = 9

Now,

(x - 1/x)3 = 93

x3 - 1/x3 - 3*x*1/x (x - 1/x) = 729

x3 - 1/x3 - 3*x*1/x (x - 1/x) = 729

now substituting the value of x - 1/x

x3 - 1/x3 - 3*9 = 729

x3 - 1/x3 - 27 = 729

x3 - 1/x3 = 729 + 27

x3 - 1/x3 = 756

• 10

x2 + 1/x= 83

x+ 1/x2 - 2 = 83 - 2

(x - 1/x)2 = 81

(x - 1/x)= 92

x - 1/x = 9

Now,

(x - 1/x)3 = 93

x3 - 1/x3 - 3*x*1/x (x - 1/x) = 729

x3 - 1/x3 - 3*x*1/x (x - 1/x) = 729

now substituting the value of x - 1/x

x3 - 1/x3 - 3*9 = 729

x3 - 1/x3 - 27 = 729

x3 - 1/x3 = 729 + 27

x3 - 1/x3 = 756

• 7
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