Question

How do you express $\dfrac{1}{{{x^6} - {x^3}}}$ in partial fractions?

Answer 1

Hint: This problem deals with reducing the given complex fraction into partial fractions. In algebra, the partial fraction decomposition or partial fraction expansion of a rational fraction is an operation that consists of expressing the fraction as a sum of a polynomial and one or several fractions with a simpler denominator.

Complete step-by-step answer:
Given a complex fraction which is as given below:
\[ \Rightarrow \dfrac{1}{{{x^6} - {x^3}}}\]
Now consider the denominator of the above given partial fraction, as shown below:
$ \Rightarrow {x^6} - {x^3}$
Now this a polynomial can be written as a product of two other polynomials as shown:
$ \Rightarrow {x^6} - {x^3} = \left( {{x^3} - 1} \right){x^3}$
Now considering the given complex fraction as given below:
$ \Rightarrow \dfrac{1}{{{x^6} - {x^3}}} = \dfrac{1}{{\left( {{x^3} - 1} \right){x^3}}}$
The right hand side of the above equation can be split into:
$ \Rightarrow \dfrac{1}{{{x^6} - {x^3}}} = \dfrac{1}{{\left( {{x^3} - 1} \right)}} - \dfrac{1}{{{x^3}}}$
Now consider the polynomial ${x^3} - 1$, this can be written as shown below:
$ \Rightarrow {x^3} - 1 = \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)$
Now consider the fraction, as shown:
$ \Rightarrow \dfrac{1}{{{x^3} - 1}} = \dfrac{1}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}$
Multiply and divide the right hand side of the equation with 3:
\[ \Rightarrow \dfrac{1}{{{x^3} - 1}} = \dfrac{1}{3} \cdot \dfrac{3}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\]
Now the numerator of the right hand side of the equation can be written as:
\[ \Rightarrow \dfrac{1}{{{x^3} - 1}} = \dfrac{1}{3}\left[ {\dfrac{{\left( {{x^2} + x + 1} \right) - \left( {{x^2} + x - 2} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}} \right]\]
Now the polynomial \[{x^2} + x - 2\] is factored into $\left( {x - 1} \right)\left( {x + 2} \right)$, now substituting it as shown:
\[ \Rightarrow \dfrac{1}{{{x^3} - 1}} = \dfrac{1}{3}\left[ {\dfrac{{\left( {{x^2} + x + 1} \right) - \left( {x - 1} \right)\left( {x + 2} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}} \right]\]
Now splitting the fractions of the above equation as shown:
\[ \Rightarrow \dfrac{1}{{{x^3} - 1}} = \dfrac{1}{3}\left[ {\dfrac{{\left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \dfrac{{\left( {x - 1} \right)\left( {x + 2} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}} \right]\]
\[ \Rightarrow \dfrac{1}{{{x^3} - 1}} = \dfrac{1}{3}\left[ {\dfrac{1}{{\left( {x - 1} \right)}} - \dfrac{{\left( {x + 2} \right)}}{{\left( {{x^2} + x + 1} \right)}}} \right]\]
So this fraction is split into these partial fractions:
\[\therefore \dfrac{1}{{{x^3} - 1}} = \dfrac{1}{{3\left( {x - 1} \right)}} - \dfrac{{\left( {x + 2} \right)}}{{3\left( {{x^2} + x + 1} \right)}}\]
Now substituting the partial fractions of \[\dfrac{1}{{{x^3} - 1}}\], in the given complex fraction as shown.
$ \Rightarrow \dfrac{1}{{{x^6} - {x^3}}} = \dfrac{1}{{\left( {{x^3} - 1} \right)}} - \dfrac{1}{{{x^3}}}$

$\therefore \dfrac{1}{{{x^6} - {x^3}}} = \dfrac{1}{{3\left( {x - 1} \right)}} - \dfrac{{\left( {x + 2} \right)}}{{3\left( {{x^2} + x + 1} \right)}} - \dfrac{1}{{{x^3}}}$

Note:
Please note that partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator. Partial fractions are a way of breaking apart fractions with polynomials in them. The process of taking a rational expression and decomposing it into simpler rational expressions that we can add or subtract to get the original rational expression is called partial fraction decomposition.