Problems from Reduction of algebraic fractions to a common denominator

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

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Development:

It is enough to use the steps:

  1. Factorization:

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

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

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

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

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

  1. 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.

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

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

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

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

Solution:

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

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