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

 A: We have that for $(4+x)(4+2x)(4+3x)\neq0 $
$$\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4  \iff 18 x^3 + 165 x^2 + 378 x + 255=0$$
$$\iff 6 x^3 + 35 x^2 + 126 x + 85=0$$
and by rational root theorem we can find that $x=-\frac 5 3$ is a root and then we obtain
$$6 x^3 + 35 x^2 + 126 x + 85=(3 x + 5) (2 x^2 + 15 x + 17)$$
A: If you insist on not dealing with a cubic equation, then maybe this can work. Notice:
$$\dfrac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)} = \dfrac{(1+x)(-1-2x)(1+3x)}{(4+x)(-4-2x)(4+3x)} = \dfrac{a(b-c)(c-4)}{b(a-c)(c-1)} = 4$$
where $c = 3x+5$, $a = x+1$ and $b = x+4.$ If you expand this quadratic in $c:$
$$c^2(a-4b)+c(3ab+4b-4a) = 0$$
and this one is easy to solve.
But I must say that learning how to factor cubic if they have nice solutions by the Rational Root Theorem etc is much more useful than trying to find clever substitutions. The other answer already covers that.
A: \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}
