What is the real constants A, B, C such that 1/x^3-x= A/x + B/x-1 + C/x+1?

Answer 1

1/(x^3-x) = A/x + B/(x-1) + C/(x+1)

denominator of LHS: (x^3 - x) = x(x^2-1) = x(x+1)(x-1)

LHS is resolved into partial fractions

1 = [A(x+1)(x-1) + Bx(x+1) + Cx(x-1)]

1 = [A(x^2-1) + B(x^2+x) + C(x^2-x)]

1 = (Ax^2 - A + Bx^2 + Bx +Cx^2 - Cx)

1 = [x^2(A+B+C) + x(B - C) -A]

comparing the coefficients of x^2, x and constant term in RHS and

LHS

A = -1 -----------------eqn(a)

A+B+C = 0------------eqn(b)

B-C = 0 ---------------eqn(c)

substituting value of A in eqn(b)

-1+ B+C = 0

=> B+C = 1 --------------eqn(d)

adding (c) and (d)

2B = 1

B = 1/2

substituting value of B in eqn(c)

1/2 - C = 0

C = 1/2

so A = -1, B = 1/2 and C = 1/2

Answer 2

I assume the equation is:

1/(x^3-x) = A/x +b/(x-1) + C/(x+1)

Put right hand side ove common denominator

1/(x^3-x) = {A*(x+1)(x-1)+Bx(x+1)+Cx(x-1)}/(x^3-x)

or

1 = A(x^2-1)+B(x^2+x)+C(x^2-x)

Now the coefficents of each power of x must suim up to match what is on the right hand side of the equation:

x^2: 0 = A+B+C zero since there are no terms like x^2 on left side

x^1: 0 = B - C ----> B =C

x^0: 1 = A ---> A = 1

So now: 0 = 1 + B + C = 1 + 2B ---> B = -1/2 and thus C = -1/2

Answer 3

4 unknowns, 1 equation...can't be done.