A real-valued function \(f\) satisfies the relation \(f(x)f(y) = f(2xy + 3) + 3f(x + y) - 3f(y) + 6y\), for all real numbers \(x\) and \(y\), then the value of \(f(8)\) is

The value of $f(8)$ is $19$. Swapping $x$ and $y$ gives $f(x)-2x=f(y)-2y$, so write $f(x)=2x+c$. Comparing the coefficients of $x$ at $y=0$ gives $2c=6$. Using $c=-2$ from the relation at $x=y=0$ gives $14$, but that constant fails the coefficient equation.