Question 1209229
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Part 1) Find all possible values of ab.


(a-b)^3 
= (a-b)(a-b)^2
= (a-b)(a^2-2ab+b^2)
= a(a^2-2ab+b^2)-b(a^2-2ab+b^2)
= (a^3-2a^2b+ab^2)+(-a^2b+2ab^2-b^3)
= a^3 + (-2a^2b-a^2b) + (ab^2+2ab^2) - b^3
= a^3 - 3a^2b + 3ab^2 - b^3
= a^3-b^3-3ab(a-b)


In short,
(a-b)^3 = a^3-b^3-3ab(a-b)
You can skip over the previous paragraph if you have this formula memorized or written on a notecard. 


Then we apply the equations <font color=green>a-b = 2</font> and <font color=blue>a^3-b^3 = 8</font> to isolate ab.
So,
(a-b)^3 = a^3-b^3-3ab(a-b)
(<font color=green>a-b</font>)^3 = <font color=blue>a^3-b^3</font>-3ab(<font color=green>a-b</font>)
(<font color=green>2</font>)^3 = <font color=blue>8</font>-3ab(<font color=green>2</font>) 
8 = 8 - 6ab
-6ab = 8-8
-6ab = 0
ab = 0/(-6)
<font color=red>ab = 0</font>


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Part 2) Find all possible values of a+b.


c = a+b
c^2 = a^2+2ab+b^2
c^2 = a^2+2*0+b^2 ..... plug in ab = 0
c^2 = a^2+b^2


Use the difference of cubes factoring formula
a^3-b^3 = (a-b)(a^2+ab+b^2)
a^3-b^3 = (a-b)(a^2+0+b^2)
a^3-b^3 = (a-b)(a^2+b^2)
<font color=blue>a^3-b^3</font> = (<font color=green>a-b</font>)c^2
<font color=blue>8</font> = <font color=green>2</font>c^2
c^2 = 8/2
c^2 = 4
c = sqrt(4) or c = -sqrt(4)
c = 2 or c = -2
<font color=red>a+b = 2 or a+b = -2</font>


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Part 3) Find all possible values of a and b.



a-b = 2
a = b+2


a^3 - b^3 = 8
(b+2)^3 - b^3 = 8
(b^3+3*b^2*2+3*b*2^2+2^3) - b^3 = 8
6b^2+12b+8 = 8
6b^2+12b = 0
6b(b+2) = 0
6b = 0 or b+2 = 0
b = 0 or b = -2


If b = 0, then a = b+2 = 0+2 = 2
One ordered pair solution is <font color=red>(a,b) = (2,0)</font>


If b = -2, then a = b+2 = -2+2 = 0
The other ordered pair solution is <font color=red>(a,b) = (0,-2)</font>


Note that you can do part 3 first to determine a,b
Then it's very easy to compute ab and a+b.


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Answers:


<font color=red>ab = 0</font>
<font color=red>a+b = 2</font> or <font color=red>a+b = -2</font>
<font color=red>(a,b) = (2,0)</font> or  <font color=red>(a,b) = (0,-2)</font>
</font>