Question 1169576
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Recall that sum of cubes factoring formula is
a^3 + b^3 = (a+b)(a^2 - ab + b^2)


Note how we have (a+b) show up. If we divide both sides by (a+b), then,


(a^3 + b^3)/(a+b) = [ (a+b)(a^2 - ab + b^2) ]/(a+b)
(a^3 + b^3)/(a+b) = a^2 - ab + b^2
a^2 - ab + b^2 = (a^3 + b^3)/(a+b)
a^2 - ab + b^2 = (812)/(14)
a^2 - ab + b^2 = 58
We'll use this later. So let's call this equation 3.
Note: since we divided by (a+b), we must require that {{{a <> -b}}} to avoid dividing by zero.


Go back to a+b = 14 and square both sides
a+b = 14
(a+b)^2 = 14^2
a^2 + 2ab + b^2 = 196
Call this equation 4


So we have equation 3 and equation 4 as
a^2 - ab + b^2 = 58
a^2 + 2ab + b^2 = 196
If we subtract straight down, the a^2 and b^2 terms cancel out and go away. We end up with 
-3ab = -138
ab = (-138)/(-3)
ab = 46


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So whatever 'a' and 'b' are, they must multiply to 46.
They must also add to 14.


Consider the factorization
(x-a)(x-b)


FOIL that out to get
x^2 - ax - bx + ab
x^2 - (a + b)x + ab


The roots of 
(x-a)(x-b) = 0
are a and b


This means the roots of 
x^2 - (a + b)x + ab = 0
are also a and b
We have the roots adding to the negative of the middle coefficient; and also the roots multiplying to the last term. I'm using one of Vieta's Formulas.


Therefore, you need to solve
x^2 - 14x + 46 = 0


I'll let you take over from here. Use the quadratic formula. 
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