Question

If x^2+1/x^2=18, find the value of 3x^3+6x-3/x^3-6/x

Answer 1

Step-by-step explanation:

here is the step

then try it urself

Answer 2

Answer:

252

Step-by-step explanation:

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

(x - 1/x)^2 = 18 - 2 = 16

x - 1/x = ± 4 (take + 4)

Cube on both sides :

x^3 - 1/x^3 - 3(1)(x - 1/x) = 64

x^3 - 1/x^3 - 3(4) = 64

x^3 - 1/x^3 = 76

Therefore, for +ve values,

=> 3x^3 + 6x - 3/x^3 - 6/x

=> 3x^3 - 3/x^3 + 6x - 6/x

=> 3(x^3 - 1/x^3) + 6(x - 1/x)

=> 3(76) + 6(4)

=> 252