Largest Cubic Sum

Multiplying the second equation by two and subtracting it from the first gives us $xyz^2-2x^2yz = xyz(z-2x) = -128x+64z = 64(z-2x)$

This tells us that either $z=2x$ or $xyz=64$ . Thus we have 2 cases:

1) If $z=2x$ , then by plugging this into the third equation we get that $3xy^2z=128x-128x=0$ , meaning that either $x=z=0$ or $y=0$ . In both these cases we see by inspection that the solutions will be $x,y,z=0$ . Hence we get $x^3+y^3+z^3=0$ .

2) If $xyz=64$ , we substitute this into each of the three equations to get $z=-y-2x$ $2x=-y-z$ $3y=2x-z$ . Subtracting the last two equation yields $2x=y$ . Next, we plug this into the first equation to get that $z=-4x$ . Hence, we arrive at $xyz=x\frac {y}{2} (-4x)=64$ , which yields $x=-2, y=-4, z=8$ . This gives us that $x^3+y^3+z^3=440$ , which is clearly the bigger solution.

One obvious solution is (0,0,0). If any of x, y, z is 0, it's easy to prove that the remaining two must be zero as well. So the only working solution with any or some of them being 0 is (0,0,0)

The only remaining possibility we need to explore is when none of them is 0. By eliminating y from the first and 2nd equation by multiplying 2nd equation with 2 before substracting with the first equation, we get: 2x^2 y z - xyz^2 = -64y - 64z +64y + 128 x = 128x-64z 2x^2 y z - xyz^2 = 128x-64z (2x-z)(xyz) = (2x-z)(64) 2x=z or xyz=64 However, if 2x=z, the third equation becomes 3xy^2z = 64(2x-z) = 0 which implies at least one of the three is zero. We no longer look at this possibility as it's been mentioned that the only answer here is (0,0,0)

The only one remaining is if xyz = 64 that makes the equations become 64z = -64y - 128x -->> z= -y-2x (i) 64x = -32y-32z -->> 2x= -y-z (ii) 192 y = 128x - 64z -->> 3y = 2x-z (iii) eliminating z using equation ii and iii, we get 2x-3y= -y-2x 4x=2y y=2x eliminating x with equation ii and ii, we get 2x+3y=2x-y-2z 4y = -2z z = -2y, as y=2x, z= -4x

xyz=64 x(2x)(-4x)=64 x^3= -8 x=-2 y=-4 z=8 we need to check back to the original equation with this result. It fits.

x^3+y^3+z^3 = 440

as there are only two solutions, we only need to compare it with 0^3+0^3+0^3

$xyz^2 = -64y - 128x ......(1)$
$x^2yz = -32y - 32z .........(2)$
$3xy^2z = 128x - 64z .......(3)$
Multiply $(2)$ with $-2$ and we get $-2x^2yz = 64y + 64z ......(4)$
$(1)+(3)+(4)$ gives us $xyz^2 - 2x^2yz + 3xy^2z = 0$
$xyz(z - 2x + 3y)$ = 0
There are 4 possibilities, they are $x=0$ , $y=0$ , $z=0$ , $z = 2x-3y$
Case 1: $x=0$
From $(1)$ , we get that $-64y = 0$ which implies that $y = 0$
From $(2)$ , we get that $-32z = 0$ which implies that $z = 0$
From case 1, we get $x^3 + y^3 + z^3 = 0 + 0 + 0 = 0$
Case 2: $y=0$
From $(1)$ , we get that $-128x = 0$ which implies that $x = 0$
From $(2)$ , we get that $-32z = 0$ which implies that $z = 0$
From case 2, we get $x^3 + y^3 + z^3 = 0 + 0 + 0 = 0$
Case 3: $z=0$
From $(3)$ , we get that $128x = 0$ which implies that $x = 0$
From $(2)$ , we get that $-32y = 0$ which implies that $y = 0$
From case 3, we get $x^3 + y^3 + z^3 = 0 + 0 + 0 = 0$
Case 4: $z=2x-3y$ and $x,y,z \neq 0$
$(3): 3xy^2z = 128x - 64z = 128x - 64(2x-3y) = 64 \times 3y$
$xy^2z = 64y$ or $xyz = 64$
Substitute $xyz = 64$ to $(1)$
$xyz^2 = 64z = -64y - 128x$ or $z = -y - 2x$
$z = 2x-3y$
$z = -y - 2x$
From both equation above, we get $2x - 3y = -y - 2x$ or $2x = y$
Substitute $y = 2x$
$z = -y-2x = -2x-2x = -4x$
Now, we have $y = 2x$ and $z = -4x$
$xyz = 64$
$x(2x)(-4x) = 64$
$-8x^3 = 64$
$x^3 = -8$ $x = -2$ $y = 2x = 2(-2) = -4$ $z = -4x = -4(-2) = 8$ The value of $x^3 + y^3 + z^3 = (-2)^3 + (-4)^3 + 8^3 = 440$
From all cases, there are 2 different values of $x^3 + y^3 + z^3$ , but the largest possible value of $x^3 + y^3 + z^3$ is 440

eq 1+ eq 3=> 2x=3y+z substituting back=>(y+z)*z=(4y+z)(3y+z)=>3y+z=z or 3y+z=-y-z=>2y+z=0=> x:y:z=-1:-2:4=>real nonzero(x,y,z)=(-2,-4,8)

Adding the first and third equations and subtracting twice the second equation, we obtain the equation $xyz^2-2x^2yz+3xy^z=0,$ which can be rewritten as $xyz(z-2x+3y)=0.$

If $x=0,$ then from the first equation of the original system we get $y=0,$ and from the third equation we get $z=0.$ Similarly, if $y=0$ or $z=0$ we can conclude that $x=y=z=0.$ This is a solution to our system of equations, and $x^3+y^3+z^3=0.$

If $x,\ y,\ z$ are all non-zero, then $z-2x+3y=0.$ This implies that $2x-z=3y.$ Rewriting the third equation as $3xy^2z=64(2x-z),$ we get $3xy^2z=64\cdot 3y,$ so $xyz=64.$ Replacing in the first equation $xyz$ by $64$ on the left hand side, we get $64z=-64y-128x,$ or $z=-y-2x$ . Combining this with $z-2x+3y=0,$ we get $2x-3y=-y-2x,$ which implies $y=2x.$ From this, $z=-y-2x=-4x.$ So $64=xyz=-8x^3,$ thus $x=-2.$ Therefore, $y=-4,$ $z=8$ and $x^3+y^3+z^3=440.$ Because $440>0,$ this is the answer.

Making the difference between the first equation and the double of the second you get:

xyz^2-2x^2yz = -128x+64z;

xyz(z-2x) = 64(z-2x)

So if z = 2x you get x=y=z=0 doing simple passage, hence xyz = 64.

So you can replace and obtain:

z = -y-2x

3y = 2x-z

xyz=64

So:

y = 2x

z = -4x

x^3 = -8

So:

x^3 = -8

z^3 = 512

y^3 = -64

So:

x^3+y^3+z^3 = 440

xyz(z)=−64y−128x (1) xyz(x)=−32y−32z (2) 3xyz(y)=128x−64z (3)

(1)-(2) xyz(z-2x)=64(z-2x) xyz=64 ,z=2x if z=2x,x=y=z=0 if xyz=64, sub xyz=64 into (1),(2),(3) simplify it and you can get: z=-y-2x (4) 2x=-y-z (5) 3y=2x-z (6)

(5)=(6)because xyz=64 is derived from (1)and(2)

equate it and you will get: y=2x,z=-4x

x^3+y^3+z^3=-55x^3=440

First,