Mathematical Induction proof for a cubic equation. If $ x^3  = x +1$, prove by induction that $ x^{3n}  = a_{n}x + b_n + \frac {c_n}{x}$, where $a_1=1, b_1=1, c_1=0$ and 
$a_n =  a_{n-1}  + b_{n-1}, b_n = a_{n-1} + b_{n-1} + c_{n-1}, c_n = a_{n-1} + c_{n-1}$ for $n=2,3,\dots $
For $n=1$ we have $x_3 = a_1x + b_1 + \frac {c_1}{x}$  = x + 1 + $\frac{0}{x}$ = x + 1, which is true.
Assume the case for $n=k$ is true, so $x^{3k} = a_kx + b_k + \frac{c_k}{x}$
So for $n = k+1$ (and using $a_n =  a_{n-1}  + b_{n-1}, b_n = a_{n-1} + b_{n-1} + c_{n-1}, c_n = a_{n-1} + c_{n-1}$ for $n=2,3,...), $ we have: 
$ \begin{align}
x^{3(k+1)} &= (a_{k+1-1}x + b_{k+1-1}){x}  + (a_{k+1-1} + b_{k+1-1}  + c_{k+1-1}) +  
( \frac{(a_{k+1-1} + c_{k+1-1})}{x}) \\
&= (a_k + b_k){x} + ( a_k + b_k + c_k) + \frac{(a_k + c_k)}{x} \\ 
&= a_{k+1}{x} + b_{k+1} + \frac{c_{k+1}}{x} \\\end{align} $
This is the same form as $ x^{3k} $ but for n=k+1. 
Therefore, if the result is true for k, it is also true for (k+1).  
Would you proceed like this or would you add the $(k+1)$th term and show that it is equal to the $k$th term with $n$ (or $k$) replaced by $k+1$?

Hint: Multiply $x^{3n}  = a_{n}x + b_n + \frac {c_n}{x}$ on the left by $x^3$ and on the right by $x+1$. Regroup and deduce a formula for $a_{n+1}$, $b_{n+1}$, $c_{n+1}$ in terms of $a_{n}$, $b_{n}$, $c_{n}$.
