SOLUTION: Let a and b be complex numbers. If a + b = 4 and a^2 + b^2 = 6 + ab, then what is a^3 + b^3?

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Question 1209398: Let a and b be complex numbers. If a + b = 4 and a^2 + b^2 = 6 + ab, then what is a^3 + b^3?
Found 2 solutions by math_tutor2020, MathTherapy:
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: 24

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 = 24

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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 = 24 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+=+%286%2Bsqrt%286%29%29%2F3 and b+=+%286-sqrt%286%29%29%2F3
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+=+%286%2Bsqrt%286%29%29%2F3 and b+=+%286-sqrt%286%29%29%2F3 in either order.

Cubing both of those produces a%5E3+=+%282%2F9%29%2A%2854%2B19%2Asqrt%286%29%29 and b%5E3+=+%282%2F9%29%2A%2854-19%2Asqrt%286%29%29
The sum of which is a%5E3%2Bb%5E3+=+%284%2F9%29%2A54+=+24

Note the sqrt%286%29 terms cancel out.

A somewhat similar question is found here

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!
Let a and b be complex numbers.  If a + b = 4 and a^2 + b^2 = 6 + ab, then what is a^3 + b^3?



a3 + b3 = (a + b)(a2 - ab + b2) ---- Applying the sum-of-cubes postulate/sum-of-cubes theorem/algebraic identity
a3 + b3 = (a + b)(a2 + b2 - ab) 
a3 + b3 = 4(6 + ab - ab) ----- Substituting 4 for a + b, and 6 + ab for a2 + b2
a3 + b3 = 4(6) = 24