Question

(x-1/x^2)/(x-1/x^3) = solve please!

Answer 1

We can reduce this not solve because there is nothing on the right-hand side.

Simplify the following:
(x-1/x^2)/(x-1/x^3)

Put each term in x-1/x^2 over the common denominator x^2:

x-1/x^2 = x^3/x^2-1/x^2:
(x^3/x^2-1/x^2)/(x-1/x^3)
x^3/x^2-1/x^2 = (x^3-1)/x^2:
((x^3-1)/x^2)/(x-1/x^3)
x^3-1 = x^3-1^3:
((x^3-1^3)/x^2)/(x-1/x^3)

Factor the difference of two cubes.

x^3-1^3 = (x-1) (x^2+x 1+1^2):
((x-1) (x^2+x+1^2)/x^2)/(x-1/x^3)
1^2 = 1:
(((x-1) (x^2+x+1))/x^2)/(x-1/x^3)

Put each term in x-1/x^3 over the common denominator x^3:

x-1/x^3 = x^4/x^3-1/x^3:
((x-1) (x^2+x+1))/(x^2 x^4/x^3-1/x^3)
x^4/x^3-1/x^3 = (x^4-1)/x^3:
((x-1) (x^2+x+1))/(x^2 (x^4-1)/x^3)
x^4-1 = (x^2)^2-1^2:
((x-1) (x^2+x+1))/(x^2 (x^2)^2-1^2/x^3)

Factor the difference of two squares.

(x^2)^2-1^2 = (x^2-1) (x^2+1):
((x-1) (x^2+x+1))/(x^2 (x^2-1) (x^2+1)/x^3)
x^2-1 = x^2-1^2:
((x-1) (x^2+x+1))/(x^2 ((x^2-1^2) (x^2+1))/x^3)

Factor the difference of two squares.

x^2-1^2 = (x-1) (x+1):
((x-1) (x^2+x+1))/(x^2 ((x-1) (x+1) (x^2+1))/x^3)

Multiply the numerator by the reciprocal of the denominator,

((x-1) (x^2+x+1))/(x^2 ((x-1) (x+1) (x^2+1))/x^3) = ((x-1) (x^2+x+1))/x^2×x^3/((x-1) (x+1) (x^2+1)):
((x-1) (x^2+x+1) x^3)/(x^2 (x-1) (x+1) (x^2+1))
((x-1) (x^2+x+1) x^3)/(x^2 (x-1) (x+1) (x^2+1)) = (-1+x)/(-1+x)×((x^2+x+1) x^3)/(x^2 (x+1) (x^2+1)) = ((x^2+x+1) x^3)/(x^2 (x+1) (x^2+1)):
((x^2+x+1) x^3)/(x^2 (x+1) (x^2+1))

Combine powers.

((x^2+x+1) x^3)/(x^2 (x+1) (x^2+1)) = ((x^2+x+1) x^(3-2))/((x+1) (x^2+1)):
((x^2+x+1) x^3-2)/((x+1) (x^2+1))
3-2 = 1:

Answer: ((x^2+x+1) x)/((x+1) (x^2+1))