1708$\textbf{Let $\mathbf{f(x) = x^{12}-x^9+x^4-x+1}$}\\\\ \textbf{Let $\mathbf{x\geq 1,}$ Then $\mathbf{f(x)=x^9(x^3-1)+x(x^3-1)+1>0}$ }\\\\ \textbf{Now for $\mathbf{x<1}$, Then $\mathbf{f(x)=x^{12}+x^4(1-x^5)+(1-x)}>0$ }\\\\
$\textbf{So Largest Interval for which $\mathbf{f(x)>0}$ is $\mathbf{x<1}$ }$\\\\ \boxed{\boxed{\mathbf{x\in\left(-\infty,1\right)}}}$
