#### Answer

a. The sums are all equal to 30.
b. The sums are all equal to 36.
c. If we choose $a=4$, $b=1$, $c=2$, the sums are all equal to 12.
d. We may inductively conclude that all the sums are all equal to $3*a$.
e. When we sum up each row, column and diagonal without replacing the variables, we can deductively conclude that all the sums are equal to $3*a$.

#### Work Step by Step

* The table is provided in the picture *
a)
$ a + b = 10 + 6 = 16 $
$ a -b-c=10-6-1=3$
$a+c=10+1=11$
Row #1: 16 , 3 , 11
$a-b+c=10-6+1=5$
$a=10$
$a+b-c=10+6-1=15$
Row #2: 5 , 10 , 15
$a-c = 10-1=9$
$a+b+c=10+6+1=17$
$a-b=10-6=4$
Row #3: 9 , 17 , 4
The result table is:
16, 3, 11
5, 10, 15
9, 17, 4
The sums are:
Row #1: $16 + 3 + 11 = 30$
Row #2: $5 + 10 + 15 = 30$
Row #3: $9 + 17 + 4 = 30$
Column #1: $16 + 5 + 9 = 30$
Column #3: $3 + 10 + 17 = 30$
Column #3: $11 + 15 + 4 = 30$
Diagonal #1: $16 + 10 + 4 = 30$
Diagonal #2: $9 + 10 + 11 = 30$
We observe that all the sums are equal to 30.
b. repeat part (a) for $a=12$, $b=5$, and $c=2$.
$ a + b = 12 + 5 = 17 $
$ a -b-c=12-5-2=5$
$a+c=12+2=14$
Row #1: 17 , 5 , 14
$a-b+c=12-5+2=9$
$a=12$
$a+b-c=12+5-2=15$
Row #2: 9 , 12 , 15
$a-c = 12-2=10$
$a+b+c=12+5+2=19$
$a-b=12-5=7$
Row #3: 10 , 19 , 7
The result table is:
17, 5, 14
9, 12, 15
10, 19, 7
The sums are:
Row #1: $17 + 5 + 14 = 36$
Row #2: $9 + 12 + 15 = 36$
Row #3: $10 + 19 + 7 = 36$
Column #1: $17 + 9 + 10 = 36$
Column #2: $5 + 12 + 19 = 36$
Column #3: $14 + 15 + 7 = 36$
Diagonal #1: $17 + 12 + 7 = 36$
Diagonal #2: $10+ 12 + 14 = 36$
We observe that all the sums are equal to 36.
c. repeat part (a) for values $a$, $b$, and $c$ of your choice.
$a = 4$, $b = 1$, $c = 2$
$ a + b = 4+1 = 5 $
$ a -b-c=4-1-2=1$
$a+c=4+2=6$
Row #1: 5 , 1 , 6
$a-b+c=4-1+2=5$
$a=4$
$a+b-c=4+1-2=3$
Row #2: 5 , 4 , 3
$a-c = 4-2=2$
$a+b+c=4+1+2=7$
$a-b=4-1=3$
Row #3: 2 , 7 , 3
The result table is:
5, 1, 6
5, 4, 3
2, 7, 3
The sums are:
Row #1: $5 +1+6 = 12$
Row #2: $5+4+3 = 12$
Row #3: $2+7+3 = 12$
Column #1: $5+5+2 = 12$
Column #2: $1+4+7 = 12$
Column #3: $6+3+3 = 12$
Diagonal #1: $5+4+3= 12$
Diagonal #2: $2+4+6= 12$
We observe that all the sums are equal to 12.
d.
We may inductively conclude that they are all equal to $a*3$
e.
If we sum up each row, column and diagonal without replacing the variables, we can get the results in variable form, which could deductively prove our conjecture.
Row #1: $a+b+a-b-c+a+c=a+a+a=3a$
Row #2: $a-b+c+a+a+b-c=a+a+a=3a$
Row #3: $a-c+a+b+c+a-b=a+a+a=3a$
Column #1: $a+b+a-b+c+a-c=a+a+a=3a$
Column #2: $a-b-c+a+a+b+c=a+a+a=3a$
Column #2: $a+c+a+b-c+a-b=a+a+a=3a$
Diagonal #1: $a+b+a+a-b=a+a+a=3a$
Diagonal #2: $a-c+a+a+c=a+a+a=3a$
We used deductive reasoning to prove that all the sums are equal to $3a$.