If x2 +ăź=27 , nnd the value of.xs__x2

1.	If x2 +ăź=27 , nnd the value of.xs__x2
Answer» x^2+ 1/x^2= 27 (x - 1/x)^2+ 2 = 27 (x - 1/x)^2 = 27 - 2 = 25 (x - 1/x)^2 = 25 (x - 1/x) = ± 5 x³- 1/x³= (x - 1/x)^3+ 3(x - 1/x) = (5)^3 + 3(5) = 125 + 15 = 140or x³- 1/x³ = (x - 1/x)^3 + 3(x - 1/x) = (-5)^3 + 3(-5) = -6125- 15 = - 140 ∴ x³- 1/x³= 140 or - 140 Like my answer if you find it useful!

Answer»

x^2+ 1/x^2= 27

(x - 1/x)^2+ 2 = 27

(x - 1/x)^2 = 27 - 2 = 25

(x - 1/x)^2 = 25

(x - 1/x) = ± 5

x³- 1/x³= (x - 1/x)^3+ 3(x - 1/x)

= (5)^3 + 3(5)

= 125 + 15

= 140or

x³- 1/x³ = (x - 1/x)^3 + 3(x - 1/x)

= (-5)^3 + 3(-5)

= -6125- 15

= - 140

∴ x³- 1/x³= 140 or - 140

Like my answer if you find it useful!

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