You still need to factor it?

$\begin{aligned} f(x) & =x^6-10x^5-10x^4-10x^3 - 10x^2 - 10x+10 \\ & = 11x^5-10x^5-10x^4-10x^3 - 10x^2 - 10x+10 \\ & = 11x^4-10x^4-10x^3 - 10x^2 - 10x+10 \\ & = 11x^3 -10x^3 - 10x^2 - 10x+10 \\ & = 11x^2 - 10x^2 - 10x+10 \\ & = 11x - 10x+10 \\ & = 11+10 \\ & =\boxed {21} \end{aligned}$

$We\ have,$ $f\left(x\right)=x^6-10x^5-10x^4-10x^3-10x^2-10x+10$ $f\left(x\right)=x^6-10\left(x^5+x^4+x^3+x^2+x\right)+10$ $As\ \left(x^5+x^4+x^3+x^2+x\right)\ is\ a\ geometric\ progression,$ $f\left(x\right)=x^6-10\left(\frac{x\left(x^5-1\right)}{\left(x-1\right)}\right)+10$ $Now\ by\ putting\ values,$ $f\left(11\right)=11^6-10\left(\frac{\left(11^6-11\right)}{\left(11-1\right)}\right)+10$ $f\left(11\right)=11^6-11^6+11+10$ $f\left(11\right)=21$ $Hence\ the\ value\ of\ f\left(11\right)=21.$

Here we get $f(x)=x^{6}-10x^{5}-10x^{4}-10x^{3}-10x^{2}-10x+10$ . Now taking like this

$f(x)=x^5(x-10)-10x^{4}-10x^{3}-10x^{2}-10x+10$ and putting values for the first bracket. Now if we continue this thing at last we will get $x+10$ and that is $11+ 10 = 21$