$\begin{cases} \begin{aligned} x\ + \ y\ +\ z \ & =12 \\ x^2+y^2+z^2 & =12 \\ x^3+y^3+z^3 & =12 \end{aligned} \end{cases}$

If $x,y,$ and $z$ satisfy the system of equations above, what is the value of $x^4+y^4+z^4?$

The answer is 1992.

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Yes , we can , but it is much longer.

ابراهيم فقرا
- 2 years, 6 months ago

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I think it's better than 'look up newtons identities and calculate it'.

Konrad Knatz
- 2 years, 5 months ago

xyz can be found by solving (x+y+z)(x^2+y^2+z^2)=144

Akshay D Naik
- 2 years, 5 months ago

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Because x+y+z=12 -> (x+y+z)^2 = 12^2 = 144 = x^2+y^2+z^2+2xy+2xz+2yz -> 132 = 2(xy+xz+yz) -> xy+xz+yz = 66

Bolan Moonward
- 2 years, 4 months ago

how/why does xy+yz+xz=66?

spim ransley
- 2 years, 5 months ago

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Substitute the givens into OPs first computation.

Patrick Delos Reyes
- 2 years, 5 months ago

Because x+y+z=12 -> (x+y+z)^2 = 12^2 = 144 = x^2+y^2+z^2+2xy+2xz+2yz -> 132 = 2(xy+xz+yz) -> xy+xz+yz = 66

Bolan Moonward
- 2 years, 4 months ago

(xy)^2 + (xz)^2 + (yz)^2 = - 924 So we work with complex numbers

Jordan Jordanov
- 2 years, 5 months ago

I solved nearly everything on my own, and made a calculation error in the very last line. This is frustrating

Chris Jacob
- 2 years, 5 months ago

o can we find the values of x, y and z that satisfy the system of equations? Just curious how a sum of three numbers can equal the sum of their squares. I can see it with cubes as numbers can turn out negative, but squares? (0 and 1 are obviously not the solution here)

Lyuboslav Karmidzhanov
- 2 years, 4 months ago

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X=2.467+5.005i, y=7.066, z=2.467-5.005i is one solution.

Robert Gallenberger
- 2 years, 4 months ago

Thanks, Robert

Lyuboslav Karmidzhanov
- 2 years, 4 months ago

[VISIBLE CONFUSION]

Ketan Sawmy
- 1 year, 10 months ago

Let $P_n = x^n + y^n+z^n$ , where $n$ is a positive integer, $S_1=x+y+z=12$ , $S_2 = xy + yz+zx$ , and $S_3 = xyz$ . Given $P_1=P_2=P_3=12$ , we need to find $P_4 = x^4+y^4+z^4$ . By Newton's sums or Newton's identities, we have:

$\begin{aligned} P_1 & = S_1 = 12 \\ P_2 & = S_1P_1 - 2S_2 = 12(12) -2S_2 = 12 & \small \color{#3D99F6} \implies S_2 = 66 \\ P_3 & = S_1P_2 - S_2P_1 + 3S_3 = 12(12) -66(12) + 3S_3 = 12 & \small \color{#3D99F6} \implies S_3 = 220 \\ P_4 & = S_1P_3 - S_2P_2 + S_3P_1 = 12(12) -66(12) + 220(12) = \boxed{1992} \end{aligned}$

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so can we find the values of x, y and z that satisfy the system of equations? Just curious how a sum of three number can equal the sum of their squares to the same number. I can see it with cubes as numbers can turn out negative, but squares? (0 and 1 are obviously not the solution here)

Lyuboslav Karmidzhanov
- 2 years, 4 months ago

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We are talking about complex numbers that is why it is possible.

Chew-Seong Cheong
- 2 years, 4 months ago

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Thanks for clarifying, Chew. I tried to visualize it by plotting each one on a 3D graph and see if they intersect and it seems that they don't, at least not in the non-imaginary plane, i.e its roots are only complex numbers

Lyuboslav Karmidzhanov
- 2 years, 4 months ago

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@Lyuboslav Karmidzhanov – Now you will never forget.

Chew-Seong Cheong
- 2 years, 4 months ago

In the line for P3 you wrote ...66(12)+2S3=12 but it has to be 66(12)+3S3=12.

byteridr .
- 2 years, 1 month ago

$\left. \begin{aligned} 144&=(x+y+z)^2\\ &=(x^2+y^2+z^2)+2(xy+xz+yz)\\ &=12+2(xy+xz+yz) \end{aligned} \right\} \Rightarrow xy+xz+yz=66\\$

$\left. \begin{aligned} 792&=(xy+xz+yz)(x+y+z)\\ &=3xyz+[x^2(y+z)+y^2(x+z)+z^2(x+y)]\\ &=3xyz+[x^2(12-x)+y^2(12-y)+z^2(12-z)]\\ &=3xyz+132 \end{aligned} \right\} \Rightarrow xyz=220\\$

$\left. \begin{aligned} 144&=(x^3+y^3+z^3)(x+y+z)\\ &=(x^4+y^4+z^4)+[xy(x^2+y^2)+xz(x^2+z^2)+yz(y^2+z^2)]\\ &=(x^4+y^4+z^4)+[xy(12-z^2)+xz(12-y^2)+yz(12-x^2)]\\ &=(x^4+y^4+z^4)+12(xy+xz+yz)-xyz(x+y+z)\\ &=(x^4+y^4+z^4)+792-2640 \end{aligned} \right\} \Rightarrow x^4+y^4+z^4=1992\\$

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My mistake was in assuming the variables were positive. I did the same calculation to get xyz = 220 and then erroneously concluded that a sum of 12 made this impossible!

Christine Shannon
- 2 years, 5 months ago

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This was what I did. I realised (x²+y²+z²)² would take a toll on my brain cells

Shree Ganesh
- 2 years, 2 months ago

In fact, the variables aren't even real numbers in this problem. If they were, we would have from x+y+z=12 that at least one of the numbers x,y and z is 4 or greater. By symmetry, we can assume x>=4 which means x^2>=16. Assuming y and z are real, we have x^2+y^2+z^2>=16 which contradicts the group of equations given. It only has complex solutions.

Tarmo Taipale
- 2 years, 5 months ago

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Can you please explain how you got it??

Eayan Biswas
- 1 year, 10 months ago

My workings isn't as short or as elegant as those that had already posted, but this was how I worked it out.

Using Vieta's formulas, where $x, y, z$ are the roots of a cubic polynomial:

$\sum x = 12$

$(x + y +z)^2 = x^2 + y^2 + z^2 + 2\sum xy \implies \sum xy = 66$

Let $S_n = x^n + y^n + z^n$ :

Then $S_3 - 12S_2+66S_1 - 3c = 0 \implies c = 220$

So the equation is $w^3 - 12w^2 + 66w - 220 = 0$

Now we can solve for $S_4 = x^4 + y^4 + z^4$ , by substituting the given equations in the question.

$S_4 - 12S_3 + 66S_2 - 220S_1 = 0$

$S_4 = 12(12) - 66(12) + 220(12) = 1992$

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Typo in last line, the 22 should be 220.

Costumed Creeper
- 2 years, 5 months ago

I also solved by this way. Good solution! :)

Dong kwan Yoo
- 2 years, 5 months ago

Yay Newton Sums! This is the approach I used.

Nathan Richardson
- 1 year, 11 months ago

$144 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) = 12 + 2(xy + xz + yz) \implies \boxed{xy + xz + yz = 66}$ .

$1728 = (x + y + z)^3 = x^3 + y^3 + z^3 + 3x^2y + 3xy^2 + 3x^2z + 6xyz + 3y^2z + 3xz^2 + 3yz^2 =$ $x^3 + y^3 + z^3 + 3(x + y + z)(xy + xz + yz) - 3xyz = 12 + 2376 - 3xyz = 2388 - 3xyz \implies \boxed{xyz = 220}$ .

$(20736 = (x + y + z)^4 = x^4 + y^4 + z^4 + (4x^3y + 4xy^3 + 4x^3z + 4y^3z + 4xz^3 + 4yz^3) +$

$(6x^2y^2 + 12x^2yz + 12xy^2z + 6x^2z^2 + 12xyz^2 + 6y^2z^2)$ .

$6(xy + xz + yz)^2 =6x^2y^2 + 12x^2yz + 12xy^2z + 6x^2z^2 + 12xyz^2 + 6y^2z^2$ .

and,

$4(x^2 + y^2 + z^2)(xy + xz + yz) = 4x^3y + 4xy^3 + 4x^3z + 4y^3z + 4xz^3 + 4yz^3 + 4(xyz)(x + y + z) \implies$

$4x^3y + 4xy^3 + 4x^3z + 4y^3z + 4xz^3 + 4yz^3 = 4(x^2 + y^2 + z^2)(xy + xz + yz) - 4(xyz)(x + y + z)$

$\implies (x + y + z)^4 = 20736 = x^4 + y^4 + z^4 + 6(66^2) + 4(12)(66) - 4(220)(12) = x^4 + y^4 + z^4 + 18744 \implies$

$\boxed{x^4 + y^4 + z^4 = 1992}$ .

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If $(P_i=x^i+y^i+z^i)$

According to Newton's identities ,

$P_4=\frac{4P_3P_1}{3}-P_2{P_1}^2+\frac{{P_1}^4}{6}+\frac{{P_2}^2}{2}$

so we get $P_4=1992.$

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Can we find x,y and z?

Mr. India
- 2 years, 6 months ago

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If you meant whether we can find the individual values of $x,y,z$ satisfying this system of equations, then yes!

By Vieta's formula , $x,y,z$ satisfy the equation $w^3 - 12w^2 + 66w - 220 = 0$ . Can you proceed from here?

Pi Han Goh
- 2 years, 6 months ago

Whats newton’s identities?

Golden Boy
- 2 years, 6 months ago

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A good way to find the sum of the same power of the polynomian roots , you should read

ابراهيم فقرا
- 2 years, 6 months ago

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Ok thanks! Also what grade is this problem?

Golden Boy
- 2 years, 6 months ago

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@Golden Boy – I learned it by myself, it will take less than 30 minutes to understand it and why it is right

ابراهيم فقرا
- 2 years, 5 months ago

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@ابراهيم فقرا – Ok thanks! If you were my age, I would’ve immediately befriended you but you’re 6 years older than me lol! Sorry I’m digressing! Thanks!

Golden Boy
- 2 years, 5 months ago

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@Golden Boy – Haha the age is just a number , welcome my new friend lol😜

ابراهيم فقرا
- 2 years, 5 months ago

Your solution is slightly incorrect. If you note $P_1 P_2^2$ is of degree 5 and the rest of the components are of degree 4. It should also be a $\frac{P_1^2 P_2}{2}$

Letting $-b=x+y+z, c=xy+yz+zx, -d=xyz$

From Newton's identities we have: $P_4 = -b P_3 -c P_2 -d P_1$ We can see that $b = -P_1, c = \frac{P_1^2-P_2}{2}, d = -\frac{P_1^3+2P_3-3P_1 P_2}{6}$

So $P_4 = P_1 P_3 - (\frac{P_1^2-P_2}{2}) P_2 +(\frac{P_1^3+2P_3-3P_1 P_2}{6})P_1$

$P_4 = \frac{4P_1 P_3}{3} - P_1^2 P_2 +\frac{P_1^4}{6}+\frac{P_2^2}{2}$

Alex Burgess
- 2 years, 6 months ago

x + y + z = 12 is the equation of a plane that gets no closer to the origin than (4,4,4).

x^2 + y^2 + z^2 = 12 is the equation of a sphere centered at the origin with radius 2*sqrt(3) ~ 3.46. How do these intersect?

Sean McCloskey
- 2 years, 5 months ago

Here a question: If x+y+z =12 and the sum of their square x^2 + y^2 + z^2 = 12 . Then why couldn't I replace the value x,y,z by x^2,y^2,z^2 to get x^4 + y^4 +z^4 = 12?

Thanh Thanh
- 2 years, 5 months ago

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Just because $x+y+z=12$ and $x^2 + y^2 + z^2=12$ are true, that doesn't mean that $x=x^2, y=y^2, z=z^2$ must be true.

If that's the case, the values of $x,y,z$ must take values 0 and/or 1 only. But that means that none of the 3 given equations can be satisfied.

Pi Han Goh
- 2 years, 5 months ago

If $x+y+z=k=x^2+y^2+z^2$ this doesn't mean thay $x=x^2,y=y^2,z=z^2$

ابراهيم فقرا
- 2 years, 5 months ago

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For example : $x=i,y=-1,z=1+i$ , 》》 $x^2+y^2+z^2=x+y+z=2i$ ,but , $x^2=-1$ where $x=i$ ,...

ابراهيم فقرا
- 2 years, 5 months ago

what are x y and z?

Julian Pollack
- 2 years, 5 months ago

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Two of them are complex , I don't remember them , you can find them by solving the equation , $z^3-12z^2+66z-220=0$

ابراهيم فقرا
- 2 years, 5 months ago

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Yeah so I have no clue what this means. Any chance you could explain it?

Julian Pollack
- 2 years, 5 months ago

Approximately x=7.066, y=2.467-5.005i, z=2.467+5.005i, interchangeable obviously.

Russell Arnott
- 2 years, 5 months ago

Can there be three numbers whose sum, sum of squares, sum of cubes, sum of exponential 4, sum of exponential 5,.... be equal and not equal to 0 or 1?

Shashwat Trivedi
- 2 years, 5 months ago

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click here , read my comment . for each of these values of $k$ the only possible values of $a,b,c$ are combinations of $0,1$ , you can find the the cubic polynomial using Newton's identities

ابراهيم فقرا
- 2 years, 5 months ago

This is very odd because this implies x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3

x^2+x^3+y^2+y^3+z^2+z^3 = 24 = 2(x^2+y^2+z^2) x^2(x+1) + y^2(y+1) + z^2(z+1) = 2x^2+2y^2+2z^2 which implies: x+1 = 2 y+1 = 2 z+1 = 2 so all of them are equal to 1. But they're obviously not all equal to 1. Whatever they are seem to break associative and communative properties. What gives?

Tre Baker
- 2 years, 5 months ago

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Just because $2x^2+2y^2+2z^2=(x+1)x^2+(y+1)y^2+(z+1)z^2)$ , this doesn't mean that $x+1=2,y+1=2,z+1=2$

For example , take $x=y=z=0$

ابراهيم فقرا
- 2 years, 5 months ago

Is there any formula for nth power also

Shailendra Singh Jakhar
- 2 years, 5 months ago

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You can read Newton's identities , this will help you a lot

ابراهيم فقرا
- 2 years, 5 months ago

We can find the solution of x^4 + y^4 + z^4 by the following steps: (x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+xz+yz), so that (xy+xz+yz) = 66, (x+y+z)^3 = x^3 + y^3 + z^3 + 3(xy+xz+yz) (x+y+z) – 3xyx , then we find :xyz = 220, (xy+xz+yz)^2 = (xy)^2 + (xz)^2 + (yz)^2 + 2xyz(x+y+z), then we can find (xy)^2 + (xz)^2 + (yz)^2 = - 924, (x^2 + y^2 + z^2)^2 = x^4 + y^4 + z^4 + 2((xy)^2 + (xz)^2 + (yz)^2), finally we can find x^4 + y^4 + z^4 = 1992