Is there any trick to multiplying this?: $(x+1)(x+2)(x+3)(x-1)(x-2)(x-3)$? The brackets are to be eliminated and the result is to be simplified as much as possible.
I have started with this: \begin{align}&(x+1)(x+2)(x+3)(x-1)(x-2)(x-3)\\&= (x^3+2x+3x+x+2+x+3+3x+2x+6)(x^3-2x-3x-x+2-x+3-3x-2x+6)\\&=... \end{align}
But the thing is, that this method takes very long and is prone to include calculation errors.
I'd be thankful for any help!
$(x+a)(x-a)=x^2-a^2$ so collect these terms to obtain a third degree polynomial in $x^2$. Solve that as normal.
${(x+1)(x+2)(x+3)(x-1)(x-2)(x-3) \\\qquad=~ (x^2-1)(x^2-4)(x^2-9) \\\qquad=~ x^{2\cdot 3}-(\phantom{1+4+9})x^{2\cdot 2}+(\phantom{4\cdot 9+4+9})x^2-1\cdot 4\cdot 9 \\ \qquad\quad\ddots}$