How do you solve $\dfrac{{{x}^{4}}-1}{{{x}^{3}}}=0$? \[\]

Hint: We multiply ${{x}^{3}}$ both sides of the given equation $\dfrac{{{x}^{4}}-1}{{{x}^{3}}}=0$ . We use the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to factorize and then equate each factor to zero by the property of multiplication to find the values of $x$. \[\]

Complete step by step answer:
We know how to clear denominators in rational equations. We multiply both sides the least common multiple of the denominators. Here in the given equation $\dfrac{{{x}^{4}}-1}{{{x}^{3}}}=0\Rightarrow \dfrac{{{x}^{4}}-1}{{{x}^{3}}}=\dfrac{0}{1}$ the least common multiple of the denominators ${{x}^{3}},1$ is ${{x}^{3}}\times 1={{x}^{3}}$. So we multiply ${{x}^{3}}$ both sides of the given equation to have;
\[\begin{align}
  & \dfrac{{{x}^{4}}-1}{{{x}^{3}}}\times {{x}^{3}}=0\times {{x}^{3}} \\
 & \Rightarrow {{x}^{4}}-1=0 \\
 & \Rightarrow {{\left( x \right)}^{2}}-{{\left( 1 \right)}^{2}}=0 \\
\end{align}\]
We use the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the left had side of the above step for $a={{x}^{2}},b=1$ to have
\[\begin{align}
  & \Rightarrow \left( {{x}^{2}}-1 \right)\left( {{x}^{2}}+1 \right)=0 \\
 & \Rightarrow \left( {{\left( x \right)}^{2}}-{{\left( 1 \right)}^{2}} \right)\left( {{x}^{2}}+1 \right)=0 \\
\end{align}\]
We again use the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the left had side of the above step for $a=x,b=1$ to have
\[\begin{align}
  & \Rightarrow \left( x+1 \right)\left( x-1 \right)\left( {{x}^{2}}+1 \right)=0 \\
 & \Rightarrow \left( x+1 \right)\left( x-1 \right)\left( {{\left( x \right)}^{2}}-\left( -1 \right) \right)=0 \\
\end{align}\]
We know that complex numbers $i=\sqrt{-1}$ and ${{i}^{2}}=-1$. So we put ${{i}^{2}}=-1$ in the above step to have;
\[\begin{align}
  & \Rightarrow \left( x+1 \right)\left( x-1 \right)\left( {{x}^{2}}+1 \right)=0 \\
 & \Rightarrow \left( x+1 \right)\left( x-1 \right)\left( {{\left( x \right)}^{2}}-{{\left( i \right)}^{2}} \right)=0 \\
\end{align}\]
We again use the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the left had side of the above step for $a=x,b=i$ to have
\[\Rightarrow \left( x+1 \right)\left( x-1 \right)\left( x+i \right)\left( x-i \right)=0\]
We know from property of multiplication that if product of some factors are zero then at least one of the factor is zero which means of $\left( x+1 \right),\left( x-1 \right),\left( x+i \right),\left( x-i \right)$ is zero. So we have;
\[\begin{align}
  & \Rightarrow x+1=0\text{ or }x-1=0\text{ or }x+i=0\text{ or }x-i=0 \\
 & \Rightarrow x=-1\text{ or }x=1\text{ or }x=-i\text{ or }x=i \\
\end{align}\]
Since the denominator cannot be zero ${{x}^{3}}\ne 0\Rightarrow x\ne 0$. Since 0 is not solution of the equation the obtained solutions are
\[x=-1,1,-i,i\]

Note: We note that if the denominator would have been ${{x}^{3}}-1$ then we have to exclude the values of $x$ such that ${{x}^{3}}-1=0$ from the set of solutions which would have been $x=-1,-i,i$. We note that obtained solutions are the fourth four roots of unity out of which $-1,1$ are real roots and $-i,i$ are complex roots.