f ( x ) = x 2 + b x g ( x ) = 9

Choice C is correct. Multiplying the given functions $f$ and $g$ yields $f\left(x\right)·g\left(x\right)=\left(x^2+bx\right)\left(9x^2-27x\right)$. Applying the distributive property to the right-hand side of this equation yields $f\left(x\right)·g\left(x\right)=\left(x^2\right)\left(9x^2-27x\right)+\left(bx\right)\left(9x^2-27x\right)$. Applying the distributive property once again to the right-hand side of this equation yields $f\left(x\right)·g\left(x\right)=\left(x^2\right)\left(9x^2\right)-\left(x^2\right)\left(27x\right)+\left(bx\right)\left(9x^2\right)-\left(bx\right)\left(27x\right)$, which is equivalent to $f\left(x\right)·g\left(x\right)=9x^4-27x^3+9b{x}^3-27b{x}^2$. Factoring out $x^3$ from the second and third terms yields $f\left(x\right)·g\left(x\right)=9x^4+\left(-27+9b\right)x^3-27b{x}^2$. Since the left-hand sides of $f\left(x\right)·g\left(x\right)=9x^4+\left(-27+9b\right)x^3-27b{x}^2$ and $f\left(x\right)·g\left(x\right)=9x^4-26x^3-3x^2$ are equal, it follows that $\left(-27+9b\right)x^3=-26x^3$, or $-27+9b=-26$, and $-27b{x}^2=-3x^2$, or $-27b=-3$. Adding $27$ to each side of $-27+9b=-26$ yields $9b=1$. Dividing each side of this equation by $9$ yields $b=\frac{1}{9}$. Similarly, dividing each side of $-27b=-3$ by $-27$ yields $b=\frac{1}{9}$. Therefore, the value of $b$ is $\frac{1}{9}$.

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.