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 .

f(x)f(y) = f(2xy + 3) + 3f(x + y) - 3f(y) + 6y

lets put x=y=0 :
$$[f(0)]^2$$ = f(3) .. equation 1.

lets  put x = 0, y = 1: 
f(0)f(1) = f(3) + 6
            = $$[f(0)]^2$$ + 6
This implies, f(1) = $$\ \dfrac{\ \left[f\left(0\right)\right]^2+6}{f\left(0\right)}$$ ...equation 2.

If we put x=1, y=0:
f(1)f(0) = f(3) + 3f(1) -3f(0)
            = $$[f(0)]^2$$ + 3f(1) -3f(0)
f(1)[f(0)-3] = f(0)[f(0)-3]
As we know from equation 2 that $$f(0)\ne f\left(1\right)$$ . Hence f(0) = 3.

Now by substituting f(0) = 3 in equation 2 .. we get the value of f(1) = 5 .
Now, f(0) =3 , so f(3) = $$[f(0)]^2$$ = 9

Now, by substituting x = 0, y = 2 :

f(0)f(2) = f(3) + 12 = 9 + 12 = 21.
This implies f(2) = 7 as f(0) = 3.

Therefore, the series f(0) = 3, f(1) = 5, f(2) = 7, f(3) = 9 etc... Hence outputs are increasing AP .

Hence, f(8) = 2(8) + 3 = 19.