jmc

We can perform long division. Alternatively, by the Remainder Theorem, the remainder upon division is $(-1{)}^6-3=-2.$ Thus, we can write $$\begin{array}{rl}\frac{x^6-3}{x+1}& =\frac{(x^6-1)-2}{x+1}\\ & =\frac{x^6-1}{x+1}-\frac{2}{x+1}\\ & =\frac{(x^3-1)(x^3+1)}{x+1}-\frac{2}{x+1}\\ & =\frac{(x^3-1)(x+1)(x^2-x+1)}{x+1}-\frac{2}{x+1}\\ & =(x^3-1)(x^2-x+1)-\frac{2}{x+1}\\ & =x^5-x^4+x^3-x^2+x-1-\frac{2}{x+1}.\end{array}$$Thus, the quotient is $\boxed{x^5-x^4+x^3-x^2+x-1}.$