<s>[INST] Solving $\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 $. Simply bringing it to a common denominator does not lead me to success How can I solve this equation?
$$\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 $$
Simply bringing it to a common denominator does not lead me to success
What I tried

 [/INST] \begin{equation}
\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}
=4 \\
\text{We multiply both sides by teh common denominator}\\ 
(1+x)(1+2x)(1+3x\quad =\quad 4\big((4+x)(4+2x)(4+3x)\big)  \\
\text{we expand both sides} \\
 x^3 + 11 x^2 + 6 x + 1\quad =\quad 24 x^3 + 176 x^2 + 384 x + 256\\
\text{we collect terms} \\ 
18 x^3 + 165 x^2 + 378 x + 255=0\\ 
\text{we factor} \\  
3 (3 x + 5) (2 x^2 + 15 x + 17)=0\\ \\
\text{we solve} \\  
x = -5/3
\qquad\quad\space\space \space  \approx -1.6667\\ 
x = \frac{-15 + \sqrt{89}}{4}
\quad\approx  -1.3915\\
x = \frac{-15 - \sqrt{89}}{4}
\quad\approx -6.1085
\end{equation}
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