This seems like a simple question but I couldn't find anywhere to verify
Is it true that:
$$(\mathbf{a}+\mathbf{b}+\mathbf{c})\times (\mathbf{d}+\mathbf{e}+\mathbf{f}) = (\mathbf{a}\times \mathbf{d})+(\mathbf{a}\times\mathbf{e})+(\mathbf{a}\times\mathbf{f}) + (\mathbf{b}\times \mathbf{d})+(\mathbf{b}\times\mathbf{e})+(\mathbf{b}\times\mathbf{f}) + (\mathbf{c}\times \mathbf{d})+(\mathbf{c}\times\mathbf{e})+(\mathbf{c}\times\mathbf{f}) $$
Just wondering whether the following identity is true.
This was already ansered in comments, so I'll just write it in answers from Colescu's comment. We know that the cross product is distributive, meaning $(a+b)\times c=a\times c + b \times c$ and $c×(a+b)=c×a+c×b$
We can do this a bunch of times from the left to arrive to the right side of the equation $(a+b+c)×(d+e+f)=(a×d)+(a×e)+(a×f)+(b×d)+(b×e)+(b×f)+(c×d)+(c×e)+(c×f)$
So we start with $(a+b+c)×(d+e+f)$, We'll keep using the distributive propiety until we arrive at the desired equation.
$$(a+b+c)×(d+e+f)$$ $$=(a+[b+c])×(d+e+f)$$ $$=(a×(d+e+f))+([b+c]×(d+e+f))$$ $$=(a×(d+e+f))+(b×(d+e+f))+(c×(d+e+f))$$ $$=(a×(d+[e+f]))+(b×(d+[e+f]))+(c×(d+[e+f]))$$ $$=(a×d+a×[e+f])+(b×d+b×[e+f])+(c×d+c×[e+f])$$ $$=(a×d+a×e+a×f)+(b×d+b×e+b×f)+(c×d+c×e+c×f)$$ $$=(a×d)+(a×e)+(a×f)+(b×d)+(b×e)+(b×f)+(c×d)+(c×e)+(c×f)$$