If $x^{3}$ = x + 1, then show that
$x^{3 n} = a_{n} x + b_{n} - c_{n} x^{- 1}$
Where $a_{n + 1} = a_{n} b_{n} : b_{n + 1} = a_{n} + b_{n} + c_{n}$
and $c_{n + 1} = a_{n} + c_{n}$

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Solution

For n = 1, we have
$x^{3}$ = x + 1 = 1. x + 1 + 0. $x^{- 1}$
so that $a_{1} = 1, b_{1} = 1, c_{1} = 0.$
For n = 2, $x^{6} = (x^{3})^{2} = (x + 1)^{2}$
$= x^{2}$ + 2x + 1 = $\frac{x^{3}}{x}$ + 2x + 1
$= \frac{x + 1}{x}$ + 2x + 1. = 2x + 2 + 1. $x^{- 1}$
so that $a_{2} = 2, b_{2} = 2, c_{2} = 1.$
Here $a_{1} + b_{1}$ = 1 + 1 = 2 = $a_{2} .$
and $a_{1} + c_{1}$ = 1 + 0 = $c_{2} .$
Hence the assertion is valid for n = 2.
Let us now assume that
$x^{3 m} = a_{m} x + b_{m} + c_{m} x^{- 1}$
where $a_{m + 1} = a_{m} + b_{m}; b_{m + 1} = a_{m} + b_{m} + c_{m} .$
$c_{m + 1} = a_{m} + x_{m} .$ Then
$x^{3 (m + 1)} = x^{3 m} . x^{3} = (a_{m} x + b_{m} + c_{m} x^{- 1}) (x + 1)$
$= a_{m} x^{2} + b_{m} x + c_{m} + a_{m} x + b_{m} + c_{m} x^{- 1}$
$= a_{m} . \frac{x^{3}}{x} + (a_{m} + b_{m}) x + b_{m} + c_{m} + c_{m} x^{- 1}$
$= a_{m} \frac{(x + 1)}{x} a_{m + 1} x + b_{m} + c_{m} + c_{m} x^{- 1}$
$= a_{m} + a_{m} x^{- 1} + a_{m + 1} x + b_{m} + c_{m} + c_{m} x^{- 1}$
$= a_{m + 1} x + (a_{m} + b_{m} c_{m}) + (a_{m} + c_{m}) x^{- 1}$
$= a_{m + 1} x + b_{m + 1} + c_{c + 1} x^{- 1}$
Hence assertion holds for n = m + 1
Therefore by mathematical induction, the assertion holds for all natural numbers n.

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