#### Answer

$x=\frac{1}{2}$ or $x=\pm\sqrt{\frac{1}{3}}$

#### Work Step by Step

We factor by grouping:
$7x^{3}-x+1=x^{3}+3x^{2}+x$
$7x^{3}-x+1-x^{3}-3x^{2}-x=0$
$6x^{3}-3x^{2}-2x+1=0$
$3x^{2}(2x-1)-(2x-1)=0$
$(2x-1)(3x^{2}-1)=0$
$2x-1=0$ or $3x^{2}-1=0$
$x=\frac{1}{2}$ or $x=\pm\sqrt{\frac{1}{3}}$