Determine whether the subset $\{(x_1,x_2,x_3) \in \mathbb{F}^3:x_1x_2x_3 = 0\}$ is a subspace of $\mathbb{F}^{3}$ I need help determining if the following subset is a subspace of $\mathbb{F}^{3}$:
$\{(x_{1},x_{2},x_{3}) \in \mathbb{F}^3:x_{1}x_{2}x_{3} = 0$}.
Based on what I've read in my textbook, we must test that the subset is closed under vector addition and scalar multiplication (with the zero vector included).
Let $v,w \in \mathbb{F}^3$.
If $v = (x_{1},x_{2},x_{3})$ and $w = (y_{1},y_{2},y_{3})$ where $x_{1}x_{2}x_{3} = 0$ and   $y_{1}y_{2}y_{3} = 0$
then $v + w = (x_{1}+y_{1}, x_{2}+y_{2}, x_{3}+y_{3}$) 
The product of all ordered triples of the sum.
$(x_{1}+y_{1})(x_{2}+y_{2})(x_{3}+y_{3}) = 0$
Multiplied out:
$x_{1}x_{2}x_{3}+x_{1}x_{3}y_{2}+x_{2}x_{3}y_{1}+y_{1}y_{2}x_{3}+x_{1}x_{2}y_{3}+x_{1}y_{2}y_{3}+x_{2}y_{1}y_{3}+y_{1}y_{2}y_{3} = 0$
simplifying,
$x_{1}x_{3}y_{2}+x_{2}x_{3}y_{1}+y_{1}y_{2}x_{3}+x_{1}x_{2}y_{3}+x_{1}y_{2}y_{3}+x_{2}y_{1}y_{3}=0$ 
This is about where I got stuck. My guess is that the subset is not closed under addition because the given equality is only $0$ if each individual term sums to $0$. 
I already know how to check scalar multiplication for this particular problem. 
Thanks for the help.

Hint: Consider the elements $(1, 1, 0)$ and $(0, 0, 1)$, which lie in the given subset. 
