Let $\vec{x}=(x_1,x_2,x_3,x_4,x_5)$ be a binary string of length $5$ i.e. $x_k\in\{\pm1\}$. Let $\vec{a}_1,\ldots,\vec{a}_5$ be some arbitrary 3D vectors (this is not essential for the problem, I can have any 5 independent symbols here instead of $\vec{a}$-s). I need to prove the following identity if possible with some clever manipulation without explicitly calculating the terms:
1) If $\vec{x}=(+1,+1,+1,+1,+1)$ then $$\sum_{i=1}^5x_i\vec{a}_i+\sum_{i=1}^5\sum_{j=i+1}^5\sum_{k=j+1}^5x_ix_jx_k(\vec{a}_i-\vec{a}_j+\vec{a}_k)=-\mathrm{sgn}(x_1x_2x_3x_4x_5)(\vec{a}_1-\vec{a}_2+\vec{a}_3-\vec{a}_4+\vec{a}_5)+8(\vec{a}_1+\vec{a}_{5}).$$ 2) If $\vec{x}=(\underbrace{+1,\ldots}_{p\text{ '+1'-s}},\underbrace{-1,\ldots}_{q\text{ '-1'-s}})$ i.e. the strings starts in $p$ $+1$ entries and ends in q $-1$ entries where $p+q=5$ and $p,q\geq 1$ then $$\sum_{i=1}^5x_i\vec{a}_i+\sum_{i=1}^5\sum_{j=i+1}^5\sum_{k=j+1}^5x_ix_jx_k(\vec{a}_i-\vec{a}_j+\vec{a}_k)=-\mathrm{sgn}(x_1x_2x_3x_4x_5)(\vec{a}_1-\vec{a}_2+\vec{a}_3-\vec{a}_4+\vec{a}_5)+8(\vec{a}_p-\vec{a}_{p+1}).$$ 3) If $\vec{x}=(+1,\ldots)$ but not of the form like in 1) and 2) (for example, $\vec{x}=(+1,-1,+1,+1,+1)$), then $$\sum_{i=1}^5x_i\vec{a}_i+\sum_{i=1}^5\sum_{j=i+1}^5\sum_{k=j+1}^5x_ix_jx_k(\vec{a}_i-\vec{a}_j+\vec{a}_k)=-\mathrm{sgn}(x_1x_2x_3x_4x_5)(\vec{a}_1-\vec{a}_2+\vec{a}_3-\vec{a}_4+\vec{a}_5).$$