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Use algebraic manipulation to show that for three input varibles x

\(\displaystyle \sum\)m(1,2,3,4,5,6,7) = x

So far all I have is a truth table with 8 rows because I know 2^3 = 8 (starting at

So how do we go from:

= \(\displaystyle (x1x2x3+x1x2x3+x1x2x3+x1x2x3)+ (x1x2x3+x1x2x3+x1x2x3+x1x2x3) +(x1x2x3+x1x2x3+x1x2x3+x1x2x3)\) <--

Here is the final solution. Just don't know how they got the four extra terms OR how they got there...

I am lost in the dark...

_{1}, x_{2}, and x_{3}\(\displaystyle \sum\)m(1,2,3,4,5,6,7) = x

_{1}+ x_{2}+ x_{3}So far all I have is a truth table with 8 rows because I know 2^3 = 8 (starting at

**of course, just assume the list I prepared on here is starting at***0***and***0***not**__1__- ending at**7**.)- x
_{1}x_{2}x_{3} - 0 0 0 = 0
- 0 0 1 = 1 [x! x! x] +
- 0 1 0 = 1 [x! x x!] +
- 0 1 1 = 1 [x! x x] +
- 1 0 0 = 1 [x x! x!] +
- 1 0 1 = 1 [x x! x] +
- 1 1 0 = 1 [x x x!] +
- 1 1 1 = 1 [x x x] +

So how do we go from:

= \(\displaystyle (x1x2x3+x1x2x3+x1x2x3+x1x2x3)+ (x1x2x3+x1x2x3+x1x2x3+x1x2x3) +(x1x2x3+x1x2x3+x1x2x3+x1x2x3)\) <--

**do these extra***WHERE***terms come out of nowhere??**__FOUR__Here is the final solution. Just don't know how they got the four extra terms OR how they got there...

**= x1+ x3+ x2**I am lost in the dark...

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