Now the problem is not going to be to add or subtract fractions, but now we have to go from the result to the addition and subtraction of fractions, what we mean is that now we are going to decompose a given fraction in a sum or subtraction of simpler fractions to which we know as partial fractions. Let’s go once to the examples as I think you learn better in the exercises.
Case 1. Different linear factors.
Representation of the case of different linear factors:
\cfrac{1}{(x+1)(x+2)(x+3)} = \cfrac{A}{x+1} + \cfrac{B}{x+2} + \cfrac{C}{x+3}
Example of the case of different linear factors:
Decompose \cfrac{3x^{2}-5x-52}{(x+2)(x-4)(x+5)} in its partial fractions.
Since all the factors of the denominator are linear and different, we can write the following identity:
\cfrac{3x^{2} -5x - 52}{(x + 2)(x-4)(x+5)} = \cfrac{A}{x+2} + \cfrac{B}{x-4} + \cfrac{C}{x+5}
To continue with the resolution, what we will do is eliminate all the denominators multiplying all the equality by the denominators of the first member:
(x+2)(x-4)(x+5)\left(\cfrac{3x^{2} -5x - 52}{(x + 2)(x-4)(x+5)} = \cfrac{A}{x+2} + \cfrac{B}{x-4} + \cfrac{C}{x+5}\right)
By doing the multiplication we will have the next:
3x^{2} - 5x - 52 =
\cfrac{A(x+2)(x-4)(x+5)}{x+2} + \cfrac{B(x+2)(x-4)(x+5)}{x-4} + \cfrac{C(x+2)(x-4)(x+5)}{x+5}
3x^{2} - 5x - 52 =
\cfrac{A\cancel{(x+2)}(x-4)(x+5)}{\cancel{x+2}} + \cfrac{B(x+2)\cancel{(x-4)}(x+5)}{\cancel{x-4}} +
\cfrac{C(x+2)(x-4)\cancel{(x+5)}}{\cancel{x+5}}
3x^{2} - 5x - 52 = A(x-4)(x+5) + B(x+2)(x+5) + C(x+2)(x-4)
Now what you have to proceed to do is multiply the parentheses:
3x^{2} - 5x - 52 = A(x^{2} + 5x - 4x - 20) + B(x^{2} + 5x + 2x + 10) + C(x^{2} -4x + 2x - 8)
3x^{2} - 5x - 52 = A(x^{2} + x - 20) + B(x^{2} + 7x + 10) + C(x^{2} - 2x - 8)
Once the parentheses are multiplied, we proceed to group the capital letters with respect to the terms of x^{2}, x and terms without x:
3x^{2} - 5x - 52 = (A + B + C)x^{2} + (A + 7B-2C)x -20A + 10B - 8C
Then we must match all the terms of x^{2}, x and independent terms of the second member with respect to the x^{2}, x and independent term of the first member, this is how we will obtain our equations:
\begin{array}{r c l}
3x^{2} & = & (A + B + C)x^{2} \\
-5x & = & (A + 7B - 2C)x \\
-52 & = & -20A + 10B - 8C
\end{array}We eliminate the xs and as a consequence we will have our equations to solve:
\begin{array}{r c l}
3 & = & A + B + C\\
-5 & = & A + 7B - 2C\\
-52 & = & -20A + 10B - 8C
\end{array}The solution to this system of equations is as follows:
A = \cfrac{5}{3} \qquad B =\cfrac{-4}{9} \qquad C = \cfrac{16}{9}
With these values found, we are going to replace them in our partial fractions from the beginning:
\cfrac{3x^{2} -5x - 52}{(x + 2)(x-4)(x+5)} = \cfrac{\frac{5}{3}}{x+2} + \cfrac{\frac{-4}{9}}{x-4} + \cfrac{\frac{16}{9}}{x+5}
Simplifying, we will have our next solution:
\cfrac{3x^{2} -5x - 52}{(x + 2)(x-4)(x+5)} = \cfrac{5}{3(x+2)} + \cfrac{-4}{9(x-4)} + \cfrac{16}{9(x+5)}
Case 2. Repeated linear factors.
Representation of the case of repeated linear factors:
\cfrac{1}{(x+1)(x+1)(x+2)} = \cfrac{1}{(x+1)^{2}(x+2)}=\cfrac{A}{x+1} + \cfrac{B}{(x+1)^{2}} + \cfrac{C}{x+2}
Example of the case of repeated linear factors:
Decompose the following fraction \cfrac{9x^{3}+16x^{2}+3x-10}{x^{3}(x+5)} in its partial fractions.
The process is similar to case 1, the only “difference” it has is that we will place as many capital letter coefficients as there are repeated factors, observe:
\cfrac{9x^{3} + 16x^{2} + 3x - 10}{x^{3}(x + 5)} = \cfrac{A}{x} + \cfrac{B}{x^{2}} + \cfrac{C}{x^{3}} + \cfrac{D}{x+5}
And from here we perform the same process as explained in case 1.
Let’s multiply all the equality by the denominator factors of the first member to be canceled:
x^{3}(x+5) \left(\cfrac{9x^{3} + 16x^{2} + 3x - 10}{x^{3}(x + 5)} =\cfrac{A}{x} + \cfrac{B}{x^{2}} + \cfrac{C}{x^{3}} + \cfrac{D}{x+5}\right)
We cancel terms:
9x^{3} + 16x^{2} + 3x - 10 = \cfrac{Ax^{\cancel{3}}(x+5)}{\cancel{x}} + \cfrac{Bx^{\cancel{3}}(x+5)}{\cancel{x^{2}}} + \cfrac{C\cancel{x^{3}}(x+5)}{\cancel{x^{3}}} + \cfrac{Dx^{3}\cancel{(x+5)}}{\cancel{x+5}}
9x^{3} + 16x^{2} + 3x - 10 = Ax^{2}(x+5) + Bx(x+5) + C(x+5) + Dx^{3}
We multiply the parentheses:
9x^{3} + 16x^{2} + 3x - 10 = A(x^{3} + 5x^{2}) + B(x^{2} + 5x) + C(x+5) + D(x^{3})
And now we order the terms with respect to the exponents of the xs:
9x^{3} + 16x^{2} + 3x - 10 = (A + D)x^{3} + (5A + B)x^{2} + (5B + C)x + 5C
Then you have to match all the terms of x^{3}, x^{2}, x and independent term of the second member with respect to x^{3}, x^{2}, x and independent term of the first member, this is how we will obtain our equations:
\begin{array}{r c l}
9x^{3} & = & (A + D)x^{3} \\
16x^{2} & = & (5A + B)x^{2} \\
3x & = & (5B + C)x \\
-10 & = & 5C
\end{array}We eliminate the xs and as a consequence we will have our equations to solve:
\begin{array}{r c l}
9 & = & A + D \\
16 & = & 5A + B \\
3 & = & 5B + C \\
-10 & = & 5C
\end{array}When you solve the system of equations you will have the following values:
A = 3 \qquad B = 1\qquad C = -2 \qquad D = 6
With these coefficients found, we are going to replace them in our partial fractions from the beginning:
\cfrac{9x^{3} + 16x^{2} + 3x - 10}{x^{3}(x + 5)} = \cfrac{A}{x} + \cfrac{B}{x^{2}} + \cfrac{C}{x^{3}} + \cfrac{D}{x+5}
\cfrac{3}{x} + \cfrac{1}{x^{2}} - \cfrac{2}{x^{3}} + \cfrac{6}{x+5}
That means that our partial fractions are:
\cfrac{3}{x} + \cfrac{1}{x^{2}} - \cfrac{2}{x^{3}} + \cfrac{6}{x+5}
Case 3. Different quadratic factors.
Representation of the case of different quadratic factors:
\cfrac{1}{(x^{2} - x + 1)(x^{2} - x + 2)} = \cfrac{Ax + B}{x^{2} - x + 1} + \cfrac{Cx + D}{x^{2} - x + 2}
Example of the case of different quadratic factors:
When fractions are presented that have quadratic factors in the denominator, the difference that there will be compared to the two previous cases is that in the numerator the term of x will appear at the beginning when we raise our partial fractions with the letters A, B, C … etc. With the example, the above will be clear.
Decompose the following fraction \cfrac{3x^{3}-9x^{2}+8x-10}{(x-3)(x^{3} - 2x^{2} - x - 6)} in its partial fractions.
Once the partial fractions are raised, the following procedure is exactly the same as in the previous two cases, but in this case you must first factor the denominator, and if you have noticed, factoring is not so common. To factor the term of (x^{3} - 2x^{2} - x - 6) we have to find the factor of a series of possible factors, those possible factors are found dividing among all the integer divisors of the independent term between the integer divisors of the coefficient of the term with the highest exponent. The divisors of the independent term are:
x = \pm 6 \qquad x = \pm 3 \qquad x = \pm 2 \qquad x = \pm 1
And the divisors of the coefficient of the term with the greatest exponent are:
x = \pm 1
By dividing each combination, we will have the following possible factors:
x = \pm 6 \qquad x = \pm 3 \qquad x = \pm 2 \qquad x = \pm 1
I already saved a little work and I’m just going to tell you that we’ll use x=3 as the factor that divides, so we’ll have:
\cfrac{(x^{3} - 2x^{2} - x - 6)}{(x-3)} = x^{2} + x + 2
(x^{3} - 2x^{2} - x - 6)= (x-3)(x^{2} + x + 2)
I recommend that you do the procedure of making the divider box.
Now we can substitute in the fraction:
\cfrac{3x^{3} - 9x^{2} + 8x - 10}{(x-3)(x^{3} - 2x^{2} - x - 6)}=
\cfrac{3x^{3} - 9x^{2} + 8x - 10}{(x-3)(x-3)(x^{2} + x + 2)}=
\cfrac{3x^{3} - 9x^{2} + 8x - 10}{(x-3)^{2}(x^{2} + x + 2)}
Perfect, since we have factored that expression, we can proceed to raise our partial fractions, remember that we already saw what is done when there are repeated terms, but right now what we are going to see is what is done when there are quadratic terms. When there are quadratic terms, a sum of two uppercase letters will be placed as appropriate and one of them will have to be multiplied by x:
\cfrac{3x^{3} - 9x^{2} + 8x - 10}{(x-3)^{2}(x^{2} + x + 2)} = \cfrac{A}{x - 3} + \cfrac{B}{(x - 3)^{2}} + \cfrac{Cx + D}{x^{2} + x + 2}
From here the same steps that we have made in the two previous cases are made, we multiply all the equality by terms that cancel the denominator of the first member:
(x - 3)^{2}(x^{2} + x + 2)\left[\cfrac{3x^{3} - 9x^{2} + 8x - 10}{(x-3)^{2}(x^{2} + x + 2)}\right] =
(x - 3)^{2}(x^{2} + x + 2)\left[ \cfrac{A}{x - 3} + \cfrac{B}{(x - 3)^{2}} + \cfrac{Cx + D}{x^{2} + x + 2}\right]
We cancel some terms:
\cancel{(x - 3)^{2}}\cancel{(x^{2} + x + 3)}\left[\cfrac{3x^{3} - 9x^{2} + 8x - 10}{\cancel{(x-3)^{2}}\cancel{(x^{2} + x + 2)}}\right] =
\left[ \cfrac{A(x - 3)^{2}(x^{2} + x + 2)}{x - 3} + \cfrac{B(x - 3)^{2}(x^{2} + x + 3)}{(x - 3)^{2}} + \cfrac{(Cx + D)(x - 3)^{2}(x^{2} + x + 2)}{x^{2} + x + 2}\right]
We cancel more terms:
3x^{3} - 9x^{2} + 8x - 10 =
\left[ \cfrac{A(x - 3)^{\cancel{2}}(x^{2} + x + 2)}{\cancel{x - 3}} + \cfrac{B\cancel{(x - 3)^{2}}(x^{2} + x + 2)}{\cancel{(x - 3)^{2}}} + \cfrac{(Cx + D)(x - 3)^{2}\cancel{(x^{2} + x + 2)}}{\cancel{x^{2} + x + 2}}\right]
3x^{3} - 9x^{2} + 8x - 10 =A(x - 3)(x^{2} + x + 2) + B(x^{2} + x + 2) + (Cx + D)(x - 3)^{2}
Now we proceed to multiply the parentheses and reduce terms:
3x^{3} - 9x^{2} + 8x - 10 =
A(x^{3} + x^{2} + 2x - 3x^{2} - 3x - 6) + B(x^{2} +x + 2) +(Cx + D)(x^{2} - 6x + 9)
= A(x^{3} - 2x^{2} - x - 6) + B(x^{2} +x + 2) +C(x^{3} - 6x^{2} + 9x) + D(x^{2} - 6x + 9)
Now we are going to order all the terms with respect to the xs:
3x^{3} - 9x^{2} + 8x - 10 =
(A + C)x^{3} + (-2A + B - 6C + D)x^{2} + (-A + B + 9C - 6D)x - 6A + 2B + 9D
Then we must match all the terms of x^{3}, x^{2}, x and independent terms of the second member with respect to x^{3}, x^{2}, x and term independent of the first member, this is how we will obtain our equations:
\begin{array}{r c l}
3x^{3} & = & (A + C)x^{3} \\
-9x^{2} & = & (-2A + B - 6C + D)x^{2}\\
8x & = & (-A+B + 9C - 6D)x\\
-10 & = & -6A + 2B + 9D
\end{array}Eliminate the xs:
\begin{array}{r c l}
3 & = & A + C \\
-9 & = & -2A + B - 6C + D \\
8 & = & -A+B + 9C - 6D \\
-10 & = & -6A + 2B + 9D
\end{array}Solving the system of equations, we have the following values:
A = 2 \qquad B = 1 \qquad C = 1 \qquad D =0
Now we proceed to replace the values found in our partial fractions from the beginning:
\cfrac{3x^{3} - 9x^{2} + 8x - 10}{(x-3)^{2}(x^{2} + x + 2)} = \cfrac{A}{x - 3} + \cfrac{B}{(x - 3)^{2}} + \cfrac{Cx + D}{x^{2} + x + 2}
= \cfrac{2}{x - 3} + \cfrac{1}{(x - 3)^{2}} + \cfrac{x}{x^{2} + x + 2}
Finally, our partial fractions are the following:
\cfrac{3x^{3} - 9x^{2} + 8x - 10}{(x - 3)(x^{3} - 2x^{2} - x - 6)} = \cfrac{2}{x - 3} + \cfrac{1}{(x - 3)^{2}} + \cfrac{x}{x^{2} + x + 2}
Case 4. Equal quadratic factors.
Representation of the case of equal quadratic factors:
\cfrac{1}{(x^{2}-x+1)(x^{2} - x + 1)(x^{2} - x + 2)} =
\cfrac{1}{(x^{2} - x + 1)^{2}(x^{2} - x + 2)} = \cfrac{Ax+B}{x^{2} - x + 1} + \cfrac{Cx + D}{(x^{2} - x + 1)^{2}} + \cfrac{Ex + F}{x^{2} - x + 2}
Example of the case of equal quadratic factors:
Let’s make this example almost simple. Decompose \cfrac{2x^{5} + 4x^{3} - 3x^{2} + 3x - 1}{(x^{3} + 1)^{3}} in its partial fractions.
With what is observed in the denominator, it means that we have to factor using the factorization of a sum and difference of cubes, let’s work with the denominator for the moment, remember the factorization that says the following:
a^3 \pm b^3 = \left( a + b\right) \left(a^2 \pm ab + b^2 \right)
Now, our denominator must be enclosed in a sign of positive grouping so that it does not harm us in view of that exponent in the cube:
\left[ \left( x^3 + 1 \right)\right]^{3}
Now we can easily apply the factorization of the sum of cubes since the 1 does not change if we raise it to the cube:
\left[ \left( x^3 + 1\right)\right]^{3} = \left[ \left( x + 1 \right) \left(x^2 - x + 1 \right)\right]^{3}
= \left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{3}
Now that we have it factored, we can proceed to replace it in our fraction of the year:
\cfrac{2x^{5} + 4x^{3} - 3x^{2} + 3x - 1}{(x^{3} + 1)^{3}} = \cfrac{2x^{5} + 4x^{3} - 3x^{2} + 3x - 1}{\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{3}}
Let’s proceed to write our partial fractions:
\cfrac{2x^{5} + 4x^{3} - 3x^{2} + 3x - 1}{\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{3}} =
\cfrac{A}{x+1} + \cfrac{B}{(x+1)^{2}}+ \cfrac{C}{(x+1)^{3}} +
\cfrac{Dx + E}{x^{2}-x+1} + \cfrac{Fx + G}{\left(x^{2}-x+1\right)^{2}} + \cfrac{Hx + I}{\left(x^{2}-x+1\right)^{3}}
Multiply all the equality by \left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{3} so that the \left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{3} of the denominator of the first member is eliminated:
\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{3}\left[\cfrac{2x^{5} + 4x^{3} - 3x^{2} + 3x - 1}{\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{3}}\right] =
\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{3}\left[\cfrac{A}{x+1} + \cfrac{B}{(x+1)^{2}}+ \cfrac{C}{(x+1)^{3}} +\right.
\left.\cfrac{Dx + E}{x^{2}-x+1} + \cfrac{Fx + G}{\left(x^{2}-x+1\right)^{2}} + \cfrac{Hx + I}{\left(x^{2}-x+1\right)^{3}}\right]
Multiply and cancel terms:
\cancel{\left(x + 1 \right)^{3} }\cancel{\left(x^2 - x + 1 \right)^{3}}\left[\cfrac{2x^{5} + 4x^{3} - 3x^{2} + 3x - 1}{\cancel{\left(x + 1 \right)^{3}}\cancel{ \left(x^2 - x + 1 \right)^{3}}}\right] =
\cfrac{A\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{3}}{x+1} + \cfrac{B\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{3}}{(x+1)^{2}}+ \cfrac{C\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{3}}{(x+1)^{3}} +
\cfrac{(Dx + E)\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{3}}{x^{2}-x+1} + \cfrac{(Fx + G)\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{3}}{\left(x^{2}-x+1\right)^{2}} +
\cfrac{(Hx + I)\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{3}}{\left(x^{2}-x+1\right)^{3}}
Let’s cancel terms again:
\cfrac{A\left(x + 1 \right)^{\cancel{3}} \left(x^2 - x + 1 \right)^{3}}{\cancel{x+1}} + \cfrac{B\left(x + 1 \right)^{\cancel{3}} \left(x^2 - x + 1 \right)^{3}}{\cancel{(x+1)^{2}}}+ \cfrac{C\cancel{\left(x + 1 \right)^{3}} \left(x^2 - x + 1 \right)^{3}}{\cancel{(x+1)^{3}}} +
\cfrac{(Dx + E)\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{\cancel{3}}}{\cancel{x^{2}-x+1}} + \cfrac{(Fx + G)\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{\cancel{3}}}{\cancel{\left(x^{2}-x+1\right)^{2}}} +
\cfrac{(Hx + I)\left(x + 1 \right)^{3} \cancel{\left(x^2 - x + 1 \right)^{3}}}{\cancel{\left(x^{2}-x+1\right)^{3}}}
Canceled our terms we obtain the following:
A\left(x + 1 \right)^{2} \left(x^2 - x + 1 \right)^{3} + B\left(x + 1 \right) \left(x^2 - x + 1 \right)^{3}+ C\left(x^2 - x + 1 \right)^{3} +
(Dx + E)\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{2} + (Fx + G)\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right) + (Hx + I)\left(x + 1 \right)^{3}
To save a few steps of pure algebra, I will write at once the multiplications of the parentheses:
A(1-x+x^{2} + 2x^{3} - 2x^{4} + 2x^{5} + z^{6} - x^{7} + x^{8}) +
B(1 - 2x + 3x^{2} - x^{3} - x^{4} + 3x^{5} - 2x^{6} + x^{7}) +
C(1 - 3x + 6x^{2} - 7x^{3} + 6x^{4} - 3x^{5} + x^{6}) + E + Dx + Ex +
Dx^{2} + 2Ex^{3} + 2Dx^{4} + 2Ex^{4} + 2Dx^{5} + Ex^{5} + Ex^{6} + Dx^{7} + Ex^{7} + Dx^{8}+
G + Fx + 2Gx + 2Fx^{2} + Gx^{2} + Fx^{3} + Gx^{3} + Fx^{4} + 2Gx^{4} + 2Fx^{5} +
Gx^{5} + Fx^{6}+I + Hx + 3Ix + 3Hx^{2} + 3Ix^{2} + Ix^{3} + Hx^{4}
Now we are going to group terms with respect to the degree level of the exponent:
(A + D)x^{8} + (-A + B + E + D)x^{7} + (A - 2B + C + E + F) x^{6} +
(2A + 3B - 3C + 2D + 2F + G)x^{5} +
(-2A - B + 6C + 2D + 2E + F + 2G + H)x^{4} +
(2A - B - 7C + 2E + F + G + 3H)x^{3} + (A + 3B - 6C + D + 2F + G + 3H)x^{2}+
(-A - 2B - 3C + D + E + F + 2G + H + 3I)x +A + B + C + E + G + I
Then we must match all the terms of x^{8},x^{7}, x^{6}, x^{4}, x^{3}, x and independent terms of the second member with respect to the x^{8}, x^{7}, x^{6}, x^{4}, x^{3}, x and independent term of the first member, this is how we get our equations:
\begin{array}{r c l}
0 & = & (A + D)x^{8} \\
0 & = & (-A + B + E + D)x^{7} \\
0 & = & (A - 2B + C + E + F)x^{6} \\
2x^{5} & = & (2A + 3B - 3C + 2D + 2F + G)x^{5} \\
0 & = & (-2A - B + 6C + 2D + 2E + F + 2G + H)x^{4} \\
4x^{3} & = & (2A - B - 7C + 2E + F + G + 3H + I)x^{3} \\
-3x^{2} & = & (A + 3B + 6C + D + 2F + G + 3H + 3I)x^{2} \\
3x & = & (-A - 2B - 3C + D + E + F + 2G + H + 3I)x \\
-1 & = & A + B + C + E + G + I
\end{array}Eliminate the xs:
\begin{array}{r c l}
0 & = & A + D \\
0 & = & -A + B + E + D\\
0 & = & A - 2B + C + E + F \\
2 & = & 2A + 3B - 3C + 2D + 2F + G \\
0 & = & -2A - B + 6C + 2D + 2E + F + 2G + H\\
4 & = & 2A - B - 7C + 2E + F + G + 3H + I\\
-3 & = & A + 3B + 6C + D + 2F + G + 3H + 3I\\
3 & = & -A - 2B - 3C + D + E + F + 2G + H + 3I \\
-1 & = & A + B + C + E + G + I
\end{array}Solving the system of equations, we have the following values:
A = \cfrac{-7}{27} \quad B = \cfrac{-8}{27} \quad C = \cfrac{-13}{27} \quad D = \cfrac{7}{27} \quad E = \cfrac{-2}{9} \quad F = \cfrac{10}{27}
G = \cfrac{19}{27} \quad H = \cfrac{2}{9} \quad I = \cfrac{-4}{9}
Now we proceed to replace the values found in our equations at the beginning:
\cfrac{2x^{5} + 4x^{3} - 3x^{2} + 3x - 1}{\left(x + 1 \right)^{3} \left(x^2 - x + 1 \right)^{3}} =
\cfrac{\frac{-7}{27}}{x+1} + \cfrac{\frac{-8}{27}}{(x+1)^{2}}+ \cfrac{\frac{-13}{27}}{(x+1)^{3}} +
\cfrac{\frac{7}{27}x + \frac{-2}{9}}{x^{2}-x+1} + \cfrac{\frac{10}{27}x + \frac{19}{27}}{\left((x^{2}-x+1\right)^{2}} + \cfrac{\frac{2}{9}x + \frac{-4}{9}}{\left(x^{2}-x+1\right)^{2}}
We simplify:
\cfrac{-7}{27(x+1)} + \cfrac{-8}{27(x+1)^{2}}+ \cfrac{-13}{27(x+1)^{3}} +
\cfrac{\frac{7}{27}x + \frac{-6}{27}}{x^{2}-x+1} + \cfrac{10x +19}{27\left(x^{2}-x+1\right)^{2}} + \cfrac{2x - 4}{9\left(x^{2}-x+1\right)^{3}}
We simplify a little more to finally get our partial fractions:
\cfrac{2x^{5} + 4x^{3} - 3x^{2} + 3x - 1}{(x^{3} + 1)^{3}}=
\cfrac{-7}{27(x+1)} + \cfrac{-8}{27(x+1)^{2}}+ \cfrac{-13}{27(x+1)^{3}} +
\cfrac{\frac{7}{27}x + \frac{-6}{27}}{x^{2}-x+1} + \cfrac{10x +19}{27\left(x^{2}-x+1\right)^{2}} + \cfrac{2(x - 2)}{9\left(x^{2}-x+1\right)^{3}}
Now you know how to get partial fractions, go to the world for those large fractions!
Thank you for being at this moment with us:)
