6. If $x^{3}-3 \sqrt{2} x^{2}+6 x-2 \sqrt{2}-8=0$, find the value of $x^{5}-41 x^{2}+2012$.
(1 mark) If $x^{3}-3 \sqrt{2} x^{2}+6 x-2 \sqrt{2}-8=0$, find the value of $x^{5}-41 x^{2}+2012$.
6. 1998
6. Note that $8=x^{3}-3 \sqrt{2} x^{2}+6 x-2 \sqrt{2}=(x-\sqrt{2})^{3}$. It follows that $x=2+\sqrt{2}$, or $(x-2)^{2}=2$, or $x^{2}-4 x+2=0$. Thus $x^{5}-41 x^{2}+2012=\left(x^{2}-4 x+2\right)\left(x^{3}+4 x^{2}+14 x+7\right)+1998=1998$.