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Simplify: $$\frac{\left(\dfrac{3x+x^3}{1+3x^2}\right)^2-1}{\dfrac{3x^2-1}{x^3-3x}+1}\; \div\; \frac{\dfrac{9}{x^2}-\dfrac{33-x^2}{3x^2+1}}{\dfrac{3}{x^2}-\dfrac{2(x^2+3)}{(x^3-x)^2}}$$

My attempt:

$$\frac{\dfrac{(3x+x^3)(3x+x^3)-(1+3x^2)(1+3x^2)}{(1+3x^2)(1+3x^2)}}{\dfrac{3x^2-1+x^3-3x}{x^3-3x}} \div \frac{\dfrac{9(3x^2+1)-x^2(33-x^2)}{x^2(3x^2+1)}}{\dfrac{3(x^3-x)(x^3-x)-2x^2(x^2+3)}{x^2(x^3-x)(x^3-x)}} \tag1$$

$$\frac{\dfrac{(3x+x^3+1+3x^2)(3x+x^3-1-3x^2)}{(1+3x^2)(1+3x^2)}}{\dfrac{3x^2-1+x^3-3x}{x^3-3x}}\div\frac{\dfrac{9(3x^2+1)-x^2(33-x^2)}{x^2(3x^2+1)}}{\dfrac{3x^2(x^2-1)(x^2-1)-2x^2(x^2+3)}{x^2(x^3-x)(x^3-x)}} \tag2$$

$$\frac{\dfrac{(3x+x^3+1+3x^2)(3x+x^3-1-3x^2)}{(1+3x^2)(1+3x^2)}}{\dfrac{3x^2-1+x^3-3x}{x^3-3x}}\div\frac{\dfrac{9(3x^2+1)-x^2(33-x^2)}{x^2(3x^2+1)}}{\dfrac{3(x^2-1)(x^2-1)-2(x^2+3)}{x^2(x^2-1)(x^2-1)}} \tag3$$

$$\frac{\dfrac{(3x+x^3+1+3x^2)(3x+x^3-1-3x^2)}{(1+3x^2)(1+3x^2)}}{\dfrac{3x^2-1+x^3-3x}{x^3-3x}}\div \frac{\dfrac{9(3x^2+1)-x^2(33-x^2)}{(3x^2+1)}}{\dfrac{3(x^2-1)(x^2-1)-2(x^2+3)}{(x^2-1)(x^2-1)}} \tag4$$

And now I'm stuck. I don't feel I did anything wrong, but clearly there is some simplification pattern that I'm missing. The answer is: $$\frac{x(x+1)}{x^2+4x+1} $$ So obviously a lot more simplification can be done, but I can't see it. Thanks for the help.

Blue
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ronald christenkkson
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  • Have you tried writing $\left(\frac{3x+x^3}{1+3x^2}\right)^2-1=\frac{(3x+x^3-1-3x^2)(3x+x^3+1+3x^2)}{(1+3x^2)^2}$ ? Because you get two factors in the numerator which are a difference and a sum of cubes. – PinkyWay Jan 19 '23 at 14:28
  • In the interest of legibility (and perhaps your own personal sanity), I'd suggest structuring your work something like this: The expression has the form $$\frac{a}{b}\div\frac{c}{d}$$ where $a$, $b$, $c$, $d$ are appropriate sub-expressions. Then, concentrate on simplifying separately, instead of within one monster expression. This will help others follow your steps. Good luck! – Blue Jan 19 '23 at 14:59

1 Answers1

2

From $(4)$, simplify the first big fraction. Start with the numerator. Reorder all the terms before simplification. You'll get:

$\frac{(x^2 - 1)^3}{(3x^2 + 1)^2}$

Repeat for the denominator of the first big fraction. You'll get:

$\frac{(x - 1)(x^2 + 4x + 1)}{x(x^2 - 3)}$

Divide the numerator by the denominator and the first big fraction reduces to $\frac{x(x + 1)^3(x - 1)^2 (x^2 - 3)}{(3x^2 + 1)^2(x^2 + 4x + 1)}$

Repeat with the second fraction's numerator and denominator (the numerator becomes $\frac{(x^2 - 1)^2}{3x^2 + 1}$ and the denominator becomes $\frac{(3x^2 + 1)(x^2 - 3)}{(x^2 - 1)^2}$.). Dividing the numerator by the denominator means that the second big fraction becomes $\frac{(x^2 - 3)(x^2 - 1)^2}{(3x^2 + 1)^2}$.

Therefore, $(4)$ becomes $\frac{x(x + 1)^3(x - 1)^2 (x^2 - 3)}{(3x^2 + 1)^2(x^2 + 4x + 1)} \div \frac{(x^2 - 3)(x^2 - 1)^2}{(3x^2 + 1)^2}$

Flipping the second fraction gives $\frac{x(x + 1)^3(x - 1)^2 (x^2 - 3)}{(3x^2 + 1)^2(x^2 + 4x + 1)} \times \frac{(3x^2 + 1)^2}{(x^2 - 3)(x^2 - 1)^2}$.

Multiplying and simplifying gives the answer mentioned.

Yajat Shamji
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