Problems from Reduction of algebraic fractions to a common denominator

Consider the algebraic fractions $\frac{x + 3}{x^{2} - 9}$ and $\frac{x^{2}}{(x^{2} + 2) \cdot (x + 3)}$ , find two equivalent algebraic fractions with common denominator.

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Development:

It is enough to use the steps:

Factorization:

$x^{2} - 9 = (x + 3) \cdot (x - 3)$

$(x^{2} + 2) \cdot (x + 3)$

Compute the least common multiple (l.c.m). of the polynomials in the denominators:

$l c m {x^{2} - 9, (x^{2} + 2) \cdot (x + 3)} = l c m {(x - 3) \cdot (x + 3), (x^{2} + 2) \cdot (x + 3)} =$

$= (x^{2} + 2) \cdot (x + 3) \cdot (x - 3)$

We divide the l.c.m. by every denominator and multiply it by the respective numerator. The result is the numerator of the algebraic fraction; the denominator is the l.c.m.

$\frac{(x^{2} + 2) \cdot (x + 3) \cdot (x - 3)}{x^{2} - 9} = x^{2} + 2 \Rightarrow (x + 3) \cdot (x^{2} - 9) = x \cdot (x^{2} - 9) + 3 \cdot (x^{2} - 9) =$

$= x^{3} + 3 x^{2} - 9 x - 27 \Rightarrow \frac{x^{3} + 3 x^{2} - 9 x - 27}{(x^{2} + 2) \cdot (x + 3) \cdot (x - 3)}$

$\frac{(x^{2} + 2) \cdot (x + 3) \cdot (x - 3)}{(x^{2} + 2) (x + 3)} = x - 3 \Rightarrow x^{2} \cdot (x - 3) = x^{3} - 3 x^{2} \Rightarrow$

$\Rightarrow \frac{x^{3} - 3 x^{2}}{(x^{2} + 2) \cdot (x + 3) \cdot (x - 3)}$

Solution:

$\frac{x^{3} + 3 x^{2} - 9 x - 27}{(x^{2} + 2) \cdot (x + 3) \cdot (x - 3)}$ and $\frac{x^{3} - 3 x^{2}}{(x^{2} + 2) \cdot (x + 3) \cdot (x - 3)}$

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