Polynomial System with Two Equations and Three Variables

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SIMILAR PROBLEM:

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[luggepytt]

The first thing that popped into my mind was Vieta’s formula for the cubic equation, which says:

If x, y, z are solutions to t³+at²+bt+c = 0
then we know:
x+y+z = -a
xy+yz+zx = b
xyz = -c

In our case we can apply the formula backwards, and conclude that:
a = 0
b is unknown
c = -15

So x, y, z are the solutions to the equation:
t³+bt-15 = 0

In other words:
x³+bx-15 = 0
y³+by-15 = 0
z³+bz-15 = 0

It we add these three equations, we get:
(x³+y³+z³)+b(x+y+z)-45 = 0

But we know that x+y+z = 0, so what we are left with is:
(x³+y³+z³)-45 = 0

Hence:
x³+y³+z³ = 45

[MrLidless]

Fourth method:
z = -(x + y)
-xy(x + y) = 15
x³ + y³ - (x + y)³ = -3x²y - 3xy² = -3xy(x + y) = 45

[andreabaldacci]

I solved it as follows. From the first equation: x+y=-z. Squaring both sides gives x^2+y^2+2xy=z^2, or x^2+y^2=z^2-2xy. The expression to calculate can be written as (x+y)*(x^2+y^2-xy)+z^3. Substitution with what determined above gives -z*(z^2-2xy-xy)+z^3. Distributing -z and simplification lead to 3xyz, which is 45

[WahranRai]

First equation gives by multiplying separately by x^2, y^2, z^2 :
x^3 + y*x^2 + z*x^2 = 0
x*y^2 + y^3 + z*y^2 = 0
x*z^2 + y*z^2 + z^3 = 0
Add the 3 equations, factorize and utilize first equation: x^3 + y^3 + z^3 + x*y (x+y) + x*z (x+z) +y*z (y+z) = 0
x^3 + y^3 + z^3 - x*y*z - x*y*z - x*y*z= 0 ----> x^3 + y^3 + z^3 = 3*x*y*z = 45

[Okkk]

One way to find the answer to this problem is to consider the special case z=1. Then we have x+y=-1 and xy=15 from which we derive x^2+y^2=-29. Moreover, x^3+y^3 is equal to (x+y)*(x^2-xy+y^2)=44. Therefore, we obtain x^3+y^3+z^3=45.

[satyanveshi]

Method 1:
we know that
x^3+y^3+z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2 - xy -yz -zx)
Since x+y+z = 0
x^3+y^3+z^3 = 3xyz = 45 Ans

Method 2
x = -(y+z)
substituting in second equation
(y+z)(yz) = -15
make quadratic in y
y = (-z^2 +- sqrt(z^4 - 60z))/2
take z=4, y= -5/2 or -3/2(take any one other will be value of x) [we can take any value of z, z=4 gives nice results]
after substitution
x = -3/2, y = -5/2, z=4
x^3+y^3+z^4 = 45

[andy_in_colorado]

I started with subtracting z and cubing:
(x + y)^3 = -z^3
x^3 + y^3 + 3xy(x + y) = -z^3
x^3 + y^3 + z^3 = -3xy(x + y)

But x + y = -z, so:
x^3 + y^3 + z^3 = 3xyz = 45

[scottleung]

I painstakingly expanded (x+y+z)^2 and multiplied that by another x+y+z to get 18 terms altogether, and simplified it from there.

[thatkindcoder]

All the comments below are incredibly enough deriving Euler’s identity, which states that:
If x + y + z = 0, then x^3 + y^3 + z^3 = 3xyz
Thus it is 3 * 15 = 45

[tbg-brawlstars]

Just use "x³ + y³ + z³ = 3xyz" when "x+y+z =0"
Edit : Oh wait I didn't saw before that it's 2nd Method
I wrote comment in between 1st method

[Blaqjaqshellaq]

I imagined what respective values for x, y and z might fit the equations. Firstly, since the three variables have a zero sum and a positive product, it follows that two are negative and the third positive. Can you get three integers? No, because the product is 15, whose factors are 1, 3, and 5, meaning that they'd all have to be odd so their sum would be odd too and couldn't be zero. But then I figured out that {-3/2, -5/2, +4} does fit. And the sum of their cubes is -27/8 -125/8 + 64, which does equal 45 in this case. But are there other triplets?

[Unknown_User]

X+Y+Z=0
Entonces:
X=-Y-Z
Y=-Z-X
Z=-X-Y

XYZ=15
(-Y-Z)(-Z-X)(-X-Y)=15

Expandiendo la expresión:



(2XYZ +X²(Y+Z) +Y²(Z+X) +Z²(X+Y)) = -15

30 -X²(-Y-Z) -Y²(-Z-X) -Z²(-X-Y) = -15

-(X³ + Y³ + Z³) = -45

X³ + Y³ + Z³ = 45

[昆仑云路]

＝45 because x^3+y^3+z^3-3xyz can be divided by x+y+z which is zero. Then x^3+y^3+z^3-3xyz＝0.

[angelmendez-rivera]

Given the information that x, y, z are real, and that x + y + z = 0, and x·y·z = 15, and given the theorem (which is straightforward for the reader/viewer to prove) that x^3 + y^3 + z^3 – 3·x·y·z = (x + y + z)·(x^2 + y^2 + z^2 – (x·y + y·z + z·x)), one can determine that x^3 + y^3 + z^3 – 3·15 = 0·(x^2 + y^2 + z^2 – x·y – y·z – z·x), which simplifies to x^3 + y^3 + z^3 – 45 = 0. This is equivalent to x^3 + y^3 + z^3 = 45. Q. E. D.

[theblinkingbrownie]

Oh wow i multiplied 15 by 3 as a guess and it turned out right

[JimLambier]

I was able to follow the last two methods but I got lost on the first method. You mentioned using a named formula but I didn't recognize the name and have no idea how to spell it so I can't google it.

[Chimbocoto]

Is there a solution for {x, y, z}? Taking x = sqrt(15) and y = z = - (15^0.25) gets xyz =15 but x+y+z is not 0 (-0.063).

[danielbancroft]

Did anyone find a solution for X, y, z? (4, -2.5, and -1, 5 works)

[sambitkanjilal]

X^3 + y^3 + z^3 -3xyz = (x+y+z)(x^2 + y^ 2 + z^2 -xy - xz - yz) so subtract 3xyz and you get 0 then 45 follows