SOLUTION: I need to complete the conjecture based on the pattern that I see in this specific cases. Conjecture: For any number two numbers a and b, the product of (a+b) and (a-b) is always

Question 92882: I need to complete the conjecture based on the pattern that I see in this specific cases.
Conjecture: For any number two numbers a and b, the product of (a+b) and (a-b) is always equal to _?_
(2+1)*(2-1)=3=2^2-1^2 (4+2)*(4-2)=12=4^2-2^2
(3+2)*(3-2)=5=3^2-2^2 (6+3)*(6-3)=27=6^2-3^2
I have no clue on how to do this one!!
Please help!!!!!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
For any number two numbers a and b, the product of (a+b) and (a-b) is always equal to _?_
(2+1)*(2-1)=3=2^2-1^2
(4+2)*(4-2)=12=4^2-2^2
(3+2)*(3-2)=5=3^2-2^2
(6+3)*(6-3)=27=6^2-3^2
(a+b)*(a-b) = a^2-b^2
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You could check this using FOIL.
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Cheers,
Stan H.

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