`3/(x^3 + x - 2)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

You need to decompose the fraction into irreducible fractions, such that:

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

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

`3/((x-1)(x^2+ x + 2)) = A/(x-1) + (Bx+C)/(x^2+ x + 2)`

You need to bring the fractions to a common denominator:

`3 = Ax^2 + Ax + 2A + Bx^2 - Bx + Cx - C`

You need to group the terms having the same power of x:

`3 = x^2(A+B)...

You need to decompose the fraction into irreducible fractions, such that:

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

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

`3/((x-1)(x^2+ x + 2)) = A/(x-1) + (Bx+C)/(x^2+ x + 2)`

You need to bring the fractions to a common denominator:

`3 = Ax^2 + Ax + 2A + Bx^2 - Bx + Cx - C`

You need to group the terms having the same power of x:

`3 = x^2(A+B) + x(A - B + C) + 2A - C`

Comparing both sides yields:

`A+B =0 => A = -B`

A - B + C = 0 => 2A + C = 0

`2A - C = 3`

Adding the relations yields:

`4A =3 => A = 3/4 => B = -3/4 => C = -6/4`

Hence, the partial fraction decomposition is `3/((x-1)(x^2+ x + 2)) = 3/(4x-4) + (-3x-6)/(4x^2+ 4x + 8).`