Discussion
Simplify : \(\frac{x+3}{x^{3}-1} \div \frac{3x+9}{x^{2}+x+1}\)
Answer: D
The expression simplifies to \(\frac{1}{3x-3}\).
Turn the division into multiplication by the reciprocal and factorise, using \(x^{3} - 1 = (x-1)(x^{2}+x+1)\) and \(3x + 9 = 3(x+3)\):
\[ \begin{aligned} &\frac{x+3}{(x-1)(x^{2}+x+1)} \times \frac{x^{2}+x+1}{3(x+3)} \\ &= \frac{1}{3(x-1)} = \frac{1}{3x-3} \end{aligned} \]
\(3x - 3\) is the reciprocal of the answer, and the forms with \(3x + 1\) come from factorising \(x^{3} - 1\) wrongly.
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