Expand the following expression in ascending powers of $x$ as far as $x^{3}$ .
$\frac{1 - 8 x}{1 - x - 6 x^{2}}$ .

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Solution

Let $\frac{1 - 8 x}{1 - x - 6 x^{2}} = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . . . . .$

Then $(1 - 8 x) = (1 - x - 6 x^{2}) [a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . . . . .]$

On comparing coefficients, we get
$a_{0} = 1, a_{1} - a_{0} = - 8$ , whence $a_{1} = - 7$
The coefficients of higher powers of $x$ are found in succession from
the relation $a_{n} - a_{n - 1} - 6 a_{n - 2} = 0$ ;
Then, $a_{2} = - 1$ and $a_{3} = - 43$

Hence, $\frac{1 - 8 x}{1 - x - 6 x^{2}} = [1 - 7 x - x^{2} - 43 x^{3} + . . . . . .]$

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