Question

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

Answer 1

Hint: Firstly, we need to take out the common factor ${{x}^{3}}$ from the denominator to get $\dfrac{1}{{{x}^{3}}\left( {{x}^{3}}-1 \right)}$. On adding and subtracting ${{x}^{3}}$ in the numerator, the fraction will be split as \[-\dfrac{1}{{{x}^{3}}}+\dfrac{1}{\left( {{x}^{3}}-1 \right)}\]. Using the algebraic identity ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{3}}+ab+{{b}^{3}} \right)$, the fraction \[\dfrac{1}{\left( {{x}^{3}}-1 \right)}\] can be written as \[\dfrac{1}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}\]. Finally, on writing \[\dfrac{1}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}=\dfrac{a}{\left( x-1 \right)}+\dfrac{bx+c}{\left( {{x}^{2}}+x+1 \right)}\] and on equating the coefficients by comparing, we will get the values of $a$, $b$ and $c$ and hence the given fraction will be finally expressed in the partial fractions.

Complete step by step solution:
Let us write the expression given in the above question as
$\Rightarrow E=\dfrac{1}{{{x}^{6}}-{{x}^{3}}}$
Taking ${{x}^{3}}$ common in the denominator, we get
$\Rightarrow E=\dfrac{1}{{{x}^{3}}\left( {{x}^{3}}-1 \right)}$
Adding and subtracting ${{x}^{3}}$ in the numerator, we get
\[\begin{align}
  & \Rightarrow E=\dfrac{1-{{x}^{3}}+{{x}^{3}}}{{{x}^{3}}\left( {{x}^{3}}-1 \right)} \\
 & \Rightarrow E=\dfrac{1-{{x}^{3}}}{{{x}^{3}}\left( {{x}^{3}}-1 \right)}+\dfrac{{{x}^{3}}}{{{x}^{3}}\left( {{x}^{3}}-1 \right)} \\
 & \Rightarrow E=-\dfrac{1}{{{x}^{3}}}+\dfrac{1}{\left( {{x}^{3}}-1 \right)}........\left( i \right) \\
\end{align}\]
Now, let \[u=\dfrac{1}{\left( {{x}^{3}}-1 \right)}.......\left( ii \right)\].
$\Rightarrow u=\dfrac{1}{\left( {{x}^{3}}-{{1}^{3}} \right)}$
 We know that ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{3}}+ab+{{b}^{3}} \right)$. Applying this in the denominator, we get
\[\begin{align}
  & \Rightarrow u=\dfrac{1}{\left( x-1 \right)\left( {{x}^{2}}+x+{{1}^{2}} \right)} \\
 & \Rightarrow u=\dfrac{1}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)} \\
\end{align}\]
For splitting the above into partial fractions, we write
\[\Rightarrow \dfrac{1}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}=\dfrac{a}{\left( x-1 \right)}+\dfrac{bx+c}{\left( {{x}^{2}}+x+1 \right)}......\left( iii \right)\]
Taking LCM on the RHS we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}=\dfrac{a\left( {{x}^{2}}+x+1 \right)+\left( bx+c \right)\left( x-1 \right)}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)} \\
 & \Rightarrow \dfrac{1}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}=\dfrac{a{{x}^{2}}+ax+a+b{{x}^{2}}-bx+cx-c}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)} \\
 & \Rightarrow \dfrac{1}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}=\dfrac{\left( a+b \right){{x}^{2}}+\left( a-b+c \right)x+\left( a-c \right)}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)} \\
\end{align}\]
Writing the numerator on the LHS in terms of that on the RHS, we get
\[\Rightarrow \dfrac{\left( 0 \right){{x}^{2}}+\left( 0 \right)x+1}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}=\dfrac{\left( a+b \right){{x}^{2}}+\left( a-b+c \right)x+\left( a-c \right)}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}\]
On equating the coefficients of ${{x}^{2}}$, and the constant terms, we get
$\begin{align}
  & \Rightarrow a+b=0 \\
 & \Rightarrow b=-a......\left( iv \right) \\
\end{align}$
On equating the coefficients of $x$, we get
$\Rightarrow a-b+c=0$
Substituting (i) in the above equation, we get
$\begin{align}
  & \Rightarrow a-\left( -a \right)+c=0 \\
 & \Rightarrow a+a+c=0 \\
 & \Rightarrow 2a+c=0 \\
 & \Rightarrow c=-2a......\left( v \right) \\
\end{align}$
Now, equating the constant terms, we get
$\Rightarrow a-c=1$
Substituting (v) we get
$\begin{align}
  & \Rightarrow a-\left( -2a \right)=1 \\
 & \Rightarrow a+2a=1 \\
 & \Rightarrow 3a=1 \\
 & \Rightarrow a=\dfrac{1}{3}......\left( vi \right) \\
\end{align}$
Substituting (vi) in (iv) we get
$\Rightarrow b=-\dfrac{1}{3}.......\left( vii \right)$
Now, substituting (vi) in (v) we get
\[\Rightarrow c=-\dfrac{2}{3}........\left( viii \right)\]
Substituting (vi), (vii) and (viii) in (iii) we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}=\dfrac{\dfrac{1}{3}}{\left( x-1 \right)}+\dfrac{-\dfrac{1}{3}x-\dfrac{2}{3}}{\left( {{x}^{2}}+x+1 \right)} \\
 & \Rightarrow \dfrac{1}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}=\dfrac{1}{3}\left( \dfrac{1}{\left( x-1 \right)}-\dfrac{x+2}{\left( {{x}^{2}}+x+1 \right)} \right) \\
 & \Rightarrow u=\dfrac{1}{3}\left( \dfrac{1}{\left( x-1 \right)}-\dfrac{x+2}{\left( {{x}^{2}}+x+1 \right)} \right) \\
\end{align}\]
Therefore, from (i) the given expression becomes
$\begin{align}
  & \Rightarrow E=-\dfrac{1}{{{x}^{3}}}+\dfrac{1}{3}\left( \dfrac{1}{\left( x-1 \right)}-\dfrac{x+2}{\left( {{x}^{2}}+x+1 \right)} \right) \\
 & \Rightarrow E=\dfrac{1}{3\left( x-1 \right)}-\dfrac{x+2}{3\left( {{x}^{2}}+x+1 \right)}-\dfrac{1}{{{x}^{3}}} \\
\end{align}$
Hence, the given expression is expressed in partial fractions.

Note: Do not end your solution at the split form \[-\dfrac{1}{{{x}^{3}}}+\dfrac{1}{\left( {{x}^{3}}-1 \right)}\] since the fraction \[\dfrac{1}{\left( {{x}^{3}}-1 \right)}\] can be further split into the partial fractions. Whenever the denominator of a fraction can be factorized, it can be split into the partial fractions. We must note that the degree of the numerator of a partial fraction is always one less than that of the denominator.