Question 1209398
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Answer: <font color=red>24</font>


Explanation
I'll show 3 methods to solving this problem.
There could be other pathways.


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Method 1


The given equations are
a+b = 4
a^2+b^2 = 6+ab
Let's refer to these as equations (1) and (2) in the order presented.


a^3+b^3 = (a+b)*(a^2-ab+b^2) ...... sum of cubes factoring formula
a^3+b^3 = (a+b)*(a^2+b^2-ab)
a^3+b^3 = (a+b)*(6+ab-ab) ....... substitute in equation (2)
a^3+b^3 = (a+b)*(6)
a^3+b^3 = (4)*(6) ...... substitute in equation (1)
a^3+b^3 = <font color=red>24</font>


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Method 2


Square both sides of equation (1)
a+b = 4
(a+b)^2 = 4^2
a^2+2ab+b^2 = 16
(a^2+b^2)+2ab = 16
(6+ab)+2ab = 16 ..... substitute in equation (2)
6+3ab = 16
3ab = 16-6
3ab = 10
Let's call this equation (3)



Now cube both sides of equation (1)
a+b = 4
(a+b)^3 = 4^3
a^3 + 3a^2b + 3ab^2 + b^3 = 64 ... use binomial expansion formula
a^3 + b^3 + 3a^2b + 3ab^2 = 64
a^3 + b^3 + 3ab(a+b) = 64
a^3 + b^3 + 10*(4) = 64 .... substitute in equations (1) and (3)
a^3 + b^3 + 40 = 64
a^3 + b^3 = 64-40
a^3 + b^3 = <font color=red>24</font> is the final answer.


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Method 3


a+b = 4 rearranges into a = 4-b
Substitute that into a^2 + b^2 = 6 + ab and you'll get (4-b)^2 + b^2 = 6 + (4-b)*b
Expand everything out and get everything to one side.
You should arrive at 3b^2-12b+10 = 0
I'll skip steps and only provide the key milestones. 


Use of the quadratic formula yields the roots {{{b = (6+sqrt(6))/3}}} and {{{b = (6-sqrt(6))/3}}}
If b is one of those roots then a = 4-b is the other root. 
The order doesn't matter.


So you would get {{{a = (6+sqrt(6))/3}}} and {{{b = (6-sqrt(6))/3}}} in either order.


Cubing both of those produces {{{a^3 = (2/9)*(54+19*sqrt(6))}}} and {{{b^3 = (2/9)*(54-19*sqrt(6))}}}
The sum of which is {{{a^3+b^3 = (4/9)*54 = 24}}}


Note the {{{sqrt(6)}}} terms cancel out.


A somewhat similar question is found <a href="https://www.algebra.com/algebra/homework/coordinate/Linear-systems.faq.question.1209394.html">here</a>
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