Question

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

Answer 1

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

& \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{aligned}$$ Now, let $$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{aligned} & \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{aligned}$$ For splitting the above into partial fractions, we write $$\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{aligned} & \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{aligned}$$ Writing the numerator on the LHS in terms of that on the RHS, we get $$\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)}$$ On equating the coefficients of ${{x}^{2}}$, and the constant terms, we get $\begin{aligned} & \Rightarrow a+b=0 \\\ & \Rightarrow b=-a......\left( iv \right) \\\ \end{aligned}$ On equating the coefficients of $x$, we get $\Rightarrow a-b+c=0$ Substituting (i) in the above equation, we get $\begin{aligned} & \Rightarrow a-\left( -a \right)+c=0 \\\ & \Rightarrow a+a+c=0 \\\ & \Rightarrow 2a+c=0 \\\ & \Rightarrow c=-2a......\left( v \right) \\\ \end{aligned}$ Now, equating the constant terms, we get $\Rightarrow a-c=1$ Substituting (v) we get $\begin{aligned} & \Rightarrow a-\left( -2a \right)=1 \\\ & \Rightarrow a+2a=1 \\\ & \Rightarrow 3a=1 \\\ & \Rightarrow a=\dfrac{1}{3}......\left( vi \right) \\\ \end{aligned}$ Substituting (vi) in (iv) we get $\Rightarrow b=-\dfrac{1}{3}.......\left( vii \right)$ Now, substituting (vi) in (v) we get $$\Rightarrow c=-\dfrac{2}{3}........\left( viii \right)$$ Substituting (vi), (vii) and (viii) in (iii) we get $$\begin{aligned} & \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{aligned}$$ Therefore, from (i) the given expression becomes $\begin{aligned} & \Rightarrow E=-\dfrac{1}{{{x}^{3}}}+\dfrac{1}{3}\left( \dfrac{1}{\left( x-1 \right)}-\dfrac{x+2}{\left( {{x}^{2}}+x+1 \right)} \right) \\\ & \Rightarrow E=\dfrac{1}{3\left( x-1 \right)}-\dfrac{x+2}{3\left( {{x}^{2}}+x+1 \right)}-\dfrac{1}{{{x}^{3}}} \\\ \end{aligned}$ Hence, the given expression is expressed in partial fractions. **Note:** Do not end your solution at the split form $$-\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.