# The Powers that be.

Rooting through my old school books I found this old gem. The solution is pretty obvious but proving it elegantly is quite a challenge, so here goes.

Assuming none are zero, solve the following equations for a,b and c.

a + b + c = 2
a² + b² + c² = 6
a³ + b³ + c³ = 8

as usual the most elegant solution gets the cheese.
Math / Science

Last Comment
BigRat

8/22/2022 - Mon
Enabbar Ocap

Watching with interest and declining hope. My first few attempts at looking for the obvious have already failed miserably.

Those are all plus signs aren't they?
Enabbar Ocap

Aha! I have a solution. The proof is probably beyond me, apart from simple substitution to show that these numbers fit.
I can't prove that these are the only three numbers that work.
Enabbar Ocap

I'll have a stab at it.
-1, 1 and 2, not necessarily in that order.
If the numbers are any greater than that the squares will not add to 6 for the second equation.

I'll sit back now and watch how it should be done.
CompProbSolv

Thibault is on the right track.  The key observation is that no number can be greater than 2 because of the second equation.  (I'm assuming that we are dealing with integers here).

You can then see that one number MUST be 2 in order for the second equation to work.  Let's assign that to a.

Substituting this in the first equation lets you see that b=-c.  Since we know that they must be between -2 and +2 (i.e. -1, 0, or 1), the only choices are -1 and 1.

Thibault did identify the only answer, though they can be assigned to a, b, and c however you wish.

The third equation doesn't really add anything, though it does confirm that those numbers work.
BigRat

I'm assuming that we are dealing with integers here).

Why should you assume that?

The key observation is that no number can be greater than 2

Why should you assume that the answers are real?

The key observation is that the way to the solution is hidden in the title and the question.Furthermore  Thibault did identify that, for the answer, a,b and c can be assigned however you wish.
Enabbar Ocap

Why should you assume that the answers are real?

I'll have another stab (probably in the dark).

If the 'balancing' numbers weren't real then the third equation would not work.
d-glitch

Move the c terms in each equation to the RHS:
a  + b  + c  = 2   ==>   a  + b  = 2 - c      E1
a² + b² + c² = 6   ==>   a² + b² = 6 - c²   E2
a³ + b³ + c³ = 8   ==>   a³ + b³ = 8 - c³   E3

Square both sides of E1:
a² + 2ab + b² = 4 - 4c + c²                      E4

Factor both sides of E3:
(a + b)(a²  -  ab + b²) = (2 - c)(4 + 2c + c²)

Note from E1 that the linear terms are equal.
They can be canceled subject to the restriction that neither is zero.
a²  -  ab + b² = 4 + 2c + c²                      E5

Evaluate E4 - E5:
a² + 2ab + b² = 4 - 4c + c²
a² -  ab + b² = 4 + 2c + c²
----------------------------------
ab        =      - 2c                               E6

Evaluate E4 - E2, and divide each side by 2:
a² + 2ab + b² =  4 - 4c +  c²
a²            + b² =  6        -  c²
--------------------------------------
2ab        = -2 - 4c + 2c²
ab        = -1 - 2c +  c²                      E7

Evaluate E7 - E6 and solve for c:
ab      = -1 - 2c + c²
ab      =     - 2c
---------------------------
0     = -1         + c²                            E8

Substitute c = ±1 in E1:
a  + b  = 2 - (+1)
a  + b  = 1                                                E9a

a  + b  = 2 - (-1)
a  + b  = 3                                                E9b

Substitute c² = 1 in E2:
a² + b² = 6 - 1
a² + b² = 5                                               E10

By inspection E9a and E10 give two solutions:
( a, b, c) = ( +2, -1, +1)
( -1, +2, +1)

E9b and E10 give two more related solutions:
( a, b, c) = ( +2, +1, -1)
( +1, +2, -1)

The final two solutions can be found by symmetry:
( a, b, c) = ( +1, -1, +2)
( -1, +1, +2)
d-glitch

More simply with less rigor:
Move the c terms in each equation to the RHS
a  + b  + c  = 2   ==>   a  + b  = 2 - c       E1
a² + b² + c² = 6   ==>   a² + b² = 6 - c²    E2
a³ + b³ + c³ = 8   ==>   a³ + b³ = 8 - c³    E3

Set both sides of E1 to zero:
(a + b) = (2 - c) = 0

So c = +2, and a = -b is a solution for E1.
Note that it is also a solution for E3.

Substituting these values in E2:
(-b)² +  b² =   6 - (+2)²
2b² =   2
b² =   1
b  = ±1
BigRat

They can be canceled subject to the restriction that neither is zero.

Fair enough.

More simply with less rigor:........Set both sides of E1 to zero:

Actually with NO rigor.

I'm not happy about E6. ab=-2c. Since a,b and c are interchangeable and the solution is 1, -1, 2, then I can set a and b to -1 and +1 and this has to be -2c which would be -4. The reason is of course that a+b would then be zero. Your second post admits this case.

I not sure I can allow this as a solution since it, under certain conditions, admits to dividing by zero.
There is a way of solving the puzzle without doing heavy substitutional algebra.
dhsindy

The first and cubic equations are equal.  So, we only have two equations and three unknown.  So, no unique solution.

E1: a+b+c=2 divide through by 2, a/2+b/2+c/2=1

E3: a^3+b^3+c^3=8  divde through by 8, a^3 / 8+b^3 / 8+c^3 / 8=8, simplify each term yields a/2+b/2+c/2=1, same as E1.
d-glitch

There are definitely three equations in three unknowns.  And there are solutions.

The equations have a geometric interpretation in 3-space.  Each of the equations represents a geometric form, symmetric around the line x = y = z.

The first equation is a plane that includes the points:
(2, 0, 0)   (0, 2, 0)   (0, 0, 2)   (2/3, 2/3, 2/3)

The second equation is a sphere centered at (0, 0, 0) with radius equal to the square root of 6 (approx 2.449).  The intersection of E1 and E2 is a circle.

The third equation is a complex form with an spike an apex at
x = y = z = cube root of 8/3 (approx 1.387).

The intersection of E1 and E3 is an equilateral triangle with vertices at
(2, 0, 0)   (0, 2, 0)   (0, 0, 2)

One solution to this series of equations is (2, 1, -1), and all six permutations of this triplet are also solutions by symmetry, corresponding to the six points of intersection between the circle and triangle.
BigRat

I'm not happy about this approach at all. It is certainly true that the first and second equations are of a plane and a sphere. But exactly where the intersections with the third object ("complex form"?) is, is not clear to me. It would seem that one would need some sort of formular to calculate these points and that is precisely what I am looking for.

Once again, the question's title and introduction contain clues on how to solve it. I'll add an extra clue : Issac Newton.
BigRat

Now dhsindy's contribution.

The first and third equations are not equal. They are sort of similar. There is a unique solution - which is obvious and has been mentioned already - and simplyfing the cubic relation to compare it with the first equation is not mathematically correct. Simplify each term results in (a/2)³+(b/2)³+(c/2)³=1 not in a/2+b/2+c/2=1 as you suggested,
BigRat

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BigRat

New puzzle just before Christmas.
BigRat