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)   (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 and
If b is one of those roots then a = 4-b is the other root.
The order doesn't matter.

So you would get and in either order.

Cubing both of those produces and
The sum of which is

Note the terms cancel out.

A somewhat similar question is found here
Answer by MathTherapy(10552)   (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
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