$\text{ Let A curve }f\left(x\right)=\int x^{27}{\left(1+x+x^2\right)}^6\left(6x^2+5x+4\right)dx\text{ is passing through origin then }\underset{ }{f}(-1)=$

$f\left(x\right)=\int x^{27}{\left(1+x+x^2\right)}^6\left(6x^2+5x+4\right)dx$

$\begin{array}{l}=\int x^{24}{x}^3{\left(1+x+x^2\right)}^6\left(6x^2+5x+4\right)dx\\ =\int {\left(x^4\right)}^6{\left(1+x+x^2\right)}^6{x}^3\left(6x^2+5x+4\right)dx\\ =\int {\left(x^4+x^5+x^6\right)}^6\left(6x^5+5x^4+4x^3\right)dx\\ \text{ Put }{x}^6+x^5+x^4=t\\ \left(6x^5+5x^4+4x^3\right)dx=dt\\ =\int t^{6}dt=\frac{t^7}{7}+C\\ =\frac{1}{7}{\left(x^6+x^5+x^4\right)}^7=C\\ \therefore f\left(x\right)=\frac{1}{7}\left(x^6+x^5+x^4\right)+C\end{array}$

Since f(x) passing through origin
F(0)=0
$\begin{array}{l}\Rightarrow C=0\\ \therefore f\left(x\right)=\frac{1}{7}\left(x^6+x^5+x^4\right)\\ \therefore f(-1)=\frac{1}{7}(1-1+1)\\ =+\frac{1}{7}\end{array}$