If $I=\int x^{27}\left(6 x^2+5 x+4\right)\left(x^2+x+1\right)^6 d x=f(x)+C$, then $f(x)$ is equal to

$\frac{1}{7}\left(x^6+x^5+x^4\right)^7$

We have $I=\int x^{27}\left(6 x^2+5 x+4\right)\left(x^2+x+1\right)^6 d x$

$\Rightarrow I=\int\left(6 x^5+5 x^4+4 x^3\right)\left(x^6+x^5+x^4\right)^6 d x$

$\Rightarrow I=\int\left(x^6+x^5+x^4\right)^6 d\left(x^6+x^5+x^4\right) = \frac{1}{7}\left(x^6+x^5+x^4\right)^7+C$

Hence, $f(x)=\frac{1}{7}\left(x^6+x^5+x^4\right)^7$.