If $(x^3+3x)/(3x^2+1)=341/91$, then $x$ equals:

Componendo and dividendo give $(x^3+3x^2+3x+1)/(x^3-3x^2+3x-1)=(341+91)/(341-91)=432/250$, that is $(x+1{)}^3/(x-1{)}^3=216/125$. Taking cube roots, $(x+1)/(x-1)=6/5$, so $5x+5=6x-6$ and $x=11$. (6 and 5 are the two halves of that ratio, not the answer.) Check: $(1331+33)/(363+1)=1364/364=341/91$.