Question

How do you express $\dfrac{1}{{{x}^{3}}-1}$ in partial fraction?

Answer 1

Hint: We have been given an expression which we have to convert into a partial fraction. We will first reduce the polynomial expression in terms of its factors using the algebraic formula \[{{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})\]. We will then express it as the sum of a partial fraction of all of its terms using variables. Solving which, we will get the given expression in the partial fraction form.

Complete step-by-step solution:
According to the given expression, we have been given an expression having polynomials. And we have to express it in partial fraction.
We will first solve the polynomial and reduce it in terms of its factors and we will do this by making use of algebraic formula.
The expression we have is,
$\dfrac{1}{{{x}^{3}}-1}$-----(1)
To factorize the polynomial in the denominator we will use the formula: \[{{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})\].
We get,
\[{{x}^{3}}-1\]\[=\]\[(x-1)({{x}^{2}}+x+1)\]
So, we now have the expression as,
\[\dfrac{1}{{{x}^{3}}-1}=\dfrac{1}{(x-1)({{x}^{2}}+x+1)}\]-----(2)
We have the factored denominator, we will express the factors in partial fraction so we have,
\[\dfrac{1}{(x-1)({{x}^{2}}+x+1)}=\dfrac{A}{(x-1)}+\dfrac{Bx+C}{({{x}^{2}}+x+1)}\]------(3)
Taking the LCM, we get,
\[\Rightarrow \dfrac{1}{(x-1)({{x}^{2}}+x+1)}=\dfrac{A}{(x-1)}\times \dfrac{({{x}^{2}}+x+1)}{({{x}^{2}}+x+1)}+\dfrac{Bx+C}{({{x}^{2}}+x+1)}\times \dfrac{(x-1)}{(x-1)}\]
\[\Rightarrow \dfrac{1}{(x-1)({{x}^{2}}+x+1)}=\dfrac{A({{x}^{2}}+x+1)+(Bx+C)(x-1)}{(x-1)({{x}^{2}}+x+1)}\]
Multiplying the terms, we have,
\[\Rightarrow \dfrac{1}{(x-1)({{x}^{2}}+x+1)}=\dfrac{A{{x}^{2}}+Ax+A+B{{x}^{2}}+Cx-Bx-C}{(x-1)({{x}^{2}}+x+1)}\]
\[\Rightarrow \dfrac{1}{(x-1)({{x}^{2}}+x+1)}=\dfrac{(A+B){{x}^{2}}+(A-B+C)x+A-C}{(x-1)({{x}^{2}}+x+1)}\]
We will now compare the coefficients of \[{{x}^{2}}\], \[x\] and constants from the above equation and we get,
\[A+B=0\]-----(4)
\[A-B+C=0\]-----(5)
\[A-C=1\]----(6)
Now, we will add equation (4) and equation (5), we get,
\[\Rightarrow 2A+C=0\]------(7)
Adding equation (6) and equation (7), we get,
\[\Rightarrow 3A=1\]
And we get the value of \[A=\dfrac{1}{3}\].
Substituting the value of A in equation (4), we get,
\[\Rightarrow \dfrac{1}{3}+B=0\]
\[\Rightarrow B=-\dfrac{1}{3}\]
Now, we will substitute the value of A in equation (6), we get,
\[\Rightarrow \dfrac{1}{3}-C=1\]
\[\Rightarrow C=\dfrac{1}{3}-1=-\dfrac{2}{3}\]
We have obtained all the values of A, B and C, so now we will substitute these values in equation (3), we get,
\[\Rightarrow \dfrac{1}{(x-1)({{x}^{2}}+x+1)}=\dfrac{1/3}{(x-1)}+\dfrac{(-1/3)x+(-2/3)}{({{x}^{2}}+x+1)}\]
\[\Rightarrow \dfrac{1}{(x-1)({{x}^{2}}+x+1)}=\dfrac{1}{3(x-1)}-\dfrac{x+2}{3({{x}^{2}}+x+1)}\]
Therefore, we get the expression as,
\[\dfrac{1}{{{x}^{3}}-1}=\dfrac{1}{3(x-1)}-\dfrac{x+2}{3({{x}^{2}}+x+1)}\]

Note: The expression \[({{x}^{2}}+x+1)\] could not be further factorized directly so we took it as it is and expressed it as a partial fraction. The solution contains complex steps so it should be done in a neat and step wise manner. While substituting the values of A, B and C in the equation (3), the values should be substituted in the respective places and the expression should be further simplified as much as possible.