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


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
$$6 x^3 + 35 x^2 + 126 x + 85=(3 x + 5) (2 x^2 + 15 x + 17)$$
