Evaluating a Sum from an Algebraic Expression

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PLAYLISTS 🎵 :

[shmuelzehavi]

The 3 variables: x, y, z don't have to be all different. The sufficient condition for a single solution (x+y+z=0) is that 2 of them are different.

[emperorviii]

Solving this sum is simple:
((x^2)/yz) + ((y^2)/xz) + ((z^2)/xy) = 3
Now, we can look at each terms separately:
((x^2)/yz)
Multiply and divide by x.
((x^3)/xyz)

((y^2)/xz)
Multiply and divide by y.
((x^3)/xyz)

((z^2)/xy)
Multiply and divide by z.
((z^3)/xyz)

Now, if you add all the terms, you get:
((x^3) + (y^3) + (z^3))/xyz = 3

Multiply both sides by xyz
x^3 + y^3 + z^3 = 3xyz

Now we need to find x + y + z.
An important rule is that, if x + y + z = 0, x^3 + y^3 + z^3 = 3xyz.
Here, we can conclude that x + y + z = 0.

But how do we prove that if x + y + z = 0, x^3 + y^3 + z^3 = 3xyz?

Let's solve this:
x + y + z = 0
x + y = -z (1)
(x + y)^3 = (-z)^3 = -z^3 [Both -z^3 and (-z)^3 are the same, but don't apply this rule if the exponent is even.]
x^3 + y^3 + 3xy(x + y) = -z^3
Now transpose some of the terms.
x^3 + y^3 + z^3 = -3xy(x + y)
From (1), x + y = -z
x^3 + y^3 + z^3 = -3xy*-z
x^3 + y^3 + z^3 = 3xyz
Hope this has helped (although I am fully aware of the fact that most viewers of this channel do not need this explanation because to them, this equation is basic math).

[WahranRai]

Use of *Lagrange multipliers*
Lagrangian L(x, y, z, 𝛌) = x+y+z + 𝛌*(x^3 + y^3 + z^3 -3*x*y*z)
Find all the first-order partial derivatives equal to 0 :
𝛛L/𝛛x = 1 + 3λ(x^2 −y*z) = 0
𝛛L/𝛛y = 1 + 3λ(y^2 −x*z) = 0
𝛛L/𝛛z = 1 + 3λ(z^2 −x*y) = 0
𝛛L/𝛛𝛌 = x^3 + y^3 + z^3 -3*x*y*z = 0
Solve the system : solution (x= -y - z, y= y, z= z) ----> *x+ y +z = 0*

[ShefsofProblemSolving]

From the condition we get (x+y+z)((x-y)^2 + (x-z)^2 + (z-y)^2) = 0

Thus either x+y+z=0 or x=y=z. If the latter is true than the sun can be anything.

[barryzeeberg]

By inspection of the first equation, x=y=z=1 is a solution, although there may be other solutions also. So one solution to the second equation is x+y+z=3

[bettyswunghole]

Spotting the identities is the tricky part...

[davidseed]

without looking
x^3+y^3+z^3=3xyz, one solution is x=y=z, so x+y+z=3x where x can take any value.

but not in the thumbnail was the constraint that x, y, z are different.
I think actually you only need to eliminate x=y=z eg

another solution from y=z
x^3+2y^3=3xy^2, dividing by y^3 and setting w=x/y
w^3+2=3w or w^3-3w+2=0
(w-1)(w^2 +w-2)=0
(w-1)(w+2)(w-1)=0
giving w=1, ie x=y=z as before (excluded)
w=-2 x=-2y=-2z
x^2/yz + y^2/xz + z^2/xy
4 + (-1/2) +(-1/2)= 3
x+y+z= -2y+2y=0

[akifbaysal]

With a different approach came to a point where I get this eq. in terms of T=x+y+z

T^3 = 3 T (x^2 + y^2 + z^2)
T (T^2 - 3(x^2 + y^2 + z^2) = 0

therefore T=0 or T= +/- sqrt (3 .(sum of sqares x, y, z)).. I think to eliminate 2nd nonzero solution the trick with 1/2 and sum of difference squares as you did is needed.. That one is not easy to see.. İt is an essential trick.

[Qermaq]

x y and z can be the same - x=y=z = + - 1 for example.

[SuperYoonHo]

Thanks so uch sir.
And sorry...
I didn't have internet for 2 days...ㅠㅠ

[minhdoantuan]

We know xyz <> 0, so mutiply both sides by xyz, we have: x^3 + y^3 + z^3 = 3xyz
Or (x+y+z)(x^2 + y^2 + z^2 - xy - yz - xz) =0
So x+y+z = 0

[zahari]

x = y = z = 1 is obvious. So x + y + z = 3.

[giuseppemalaguti]

0... Un altra soluzione e x+y+z=sqrt(3(xy+xz+yz))