This is my solution: we have 2 cases: case1:$ x\ge 2$,from 1 we have $ y\le 4$,from 3 we have $ z\ge 6$ so from 2 we have $ y\ge 4$ so x=2,y=4,z=6 case2:$ x\le 2$,from 1 we have $ y\ge 4$,from 3 we have $ z\le 6$ so from 2 we have $ y\le 4$ so x=2,y=4,z=6 Q.e.d

tuandokim wrote: This is my solution: we have 2 cases: case1:$ x\ge 2$,from 1 we have $ y\le 4$,from 3 we have $ z\ge 6$ so from 2 we have $ y\ge 4$ so x=2,y=4,z=6 case2:$ x\le 2$,from 1 we have $ y\ge 4$,from 3 we have $ z\le 6$ so from 2 we have $ y\le 4$ so x=2,y=4,z=6 Q.e.d nice solution

Another solution could be just setting x=a+2, y=b+4, z=c+6. If one crunches the numbers, and observes the trivial inequality, he/she will find a=b=c=0, so x=2, y=4, z=6.

set $ z=3u$ and take $ x$ from the last equation into the first one, so that it turns into $ (u^3-3u)^3=3(u^3-3u)-12y+50$ and $ y^3=12y+9u-2$... take $ y=w+4$, and this system turns to $ (u^3-3u)^3-3(u^3-3u)-2=-12w$ and $ w^3+12w^2+36w=9u-18$... this can be rewritten as $ (u-2)(u^4+u^3-3u^2-2u+1)^2=-12w$ and $ w(w+6)^2=9(u-2)$... from this it's easy to conclude that $ u=2, w=0$, or equivalently, $ z=6, y=4, x=2$

Great solutions, but how did you guys get that 2, 4, 6 worked? was it just guess? Another solution involves using the cos(3*theta) formula to kill the cubic, which makes it immediately evident why 2 4 and 6 are solutions.

Rewrite the equations as the following: $ 12(4 - y) = (x - 2)(x + 1)^{2}$ $ (1)$, $ 3(z - 6) = (y - 4)(y + 2)^{2}$ $ (2)$, $ 27(x - 2) = (z - 6)(z + 3)^{2}$ $ (3)$. $ (2)$, $ (3)$ gives $ (y - 4)(x - 2)\geq{0}$ (Or $ y = - 2$, which leads to a contradiction), but from $ (1)$ we get: $ (4 - y)(x - 2)\geq{0}$. (Or $ x = - 1$, which also leads to a contradiction). So $ 0\leq{4 - y)(x - 2)\leq{0}}$. It means $ y = 4$ or $ x = 2$, but both of them lead to an unique solution: $ (x,y,z) = (2,4,6)$.

ZerAn wrote: Great solutions, but how did you guys get that 2, 4, 6 worked? was it just guess? I have the same question.

Here's a slightly more deterministic solution. Substitute $y=2r, z=3s$ to make more of the coefficients match. Then we get $x^3-3x=50-24r$ $r^3-3r=(9s-2)/8$, $s^3-3s=x$. The solution $x=r=s=2$ pretty much jumps out. We note that $x^2-3x-2=(x-2)(x+1)^2$, so $(x-2)(x+1)^2=24(2-r)$, $(r-2)(r+1)^2=9(s-2)/8$, $(s-2)(s+1)^2=(x-2)$. If any of $x,r,s$ equals 2, then they all do. If not, multiplication yields that $(x+1)^2(r+1)^2(s+1)^2=-27$, a contradiction.

Let $x=2a$, $y=4b$ and $z=6c$ (this is motivated by observing $a=b=c=1$). Then, the givens rearrange to \begin{align*} a(4a^2-3) &= 25-24b \\ b(4b^2-3) &= \frac{9c-1}{8} \\ c(4c^2-3) &= a \end{align*} Note $c < 1 \implies a \le 1 \implies 25-24b \le 1 \implies b \le 1 \implies \frac{9c-1}{8} \le 1 \implies c \ge 1$ and $c > 1 \implies a > 1 \implies 25-24b > 1 \implies b < 1 \implies \frac{9c-1}{8} < 1 \implies c < 1$. So $a=b=c=1$ is the only solution; i.e. $(x,y,z) = (2,4,6)$. ========================================================== This bounding approach is pretty motivated by the polynomial $f(x) = 4x^3-3x$, which behaves very nicely with respect to bounds; in fact, I was trying to get a solution using the identity $\cos 3x = 4\cos^3 x - 3 \cos x$ which is why I attempted the bounding in the first place. ZerAn wrote: Great solutions, but how did you guys get that 2, 4, 6 worked? was it just guess? Many substitutions later. Not realizing that this was essentially bounding, I tried a lot of different substitutiotns and one of which eventually led to solving $u-v = \tfrac{u^3-50}{3}$, $v-w = \tfrac{v^3+128}{192}$, $w-u=\tfrac{w^3}{27}$. At this point I decided to start looking for integer solutions and finally noticed $(u,v,w) = (2,16,-6)$ which gave $(x,y,z) = (2,4,6)$. In retrospect I think I should have tried the guessing first. So I guess the main difficulty of the problem was finding the triple in the first place, and afterwards it was just a matter of careful manipulations.

Answer: The only triple which satisfies our condition is $(x, y, z)=(2, 4, 6)$. (Proof) Suppose $(x-2)(y-4)(z-6) \ne 0,$ as otherwise we get $(x, y, z)=(2, 4, 6)$ as claimed. Note that the given equations may be re-written as: \[ \begin{cases} -12(y-4)=(x-2)(x-1)^2, \\ 3(z-6)=(y-4)(y+2)^2, \\ 27(x-2)=(z-6)(z+3)^2. \end{cases}\]Taking the product and cancelling the common (non-zero) factor $(x-2)(y-4)(z-6),$ we see that both sides have an opposite sign unless $(x-1)(y+2)(z+3)=0$. Evidently, the latter cannot occur as $$x=1 \Longrightarrow y=4 \Longrightarrow z=6 \Longrightarrow x=2,$$a contradiction! Similar arguments apply for the other two factors.

We claim the only solution is $(x,y,z)=\boxed{(2,4,6)}$, which can easily be checked to work. To show that this is the only solution, we have the following key lemma. Lemma: We have \begin{align*} x\ge 2 &\iff x^3-3x\ge 2 \\ y\ge 4 &\iff y^3-12y\ge 16 \\ z\ge 6 &\iff z^3-27z\ge 54, \end{align*}and similar statements with all the inequality directions reversed. Proof: The proof is ``you just check it''. $\blacksquare$ The solution to the problem is now easy. We see that \begin{align*} x\ge 2 &\iff z^3-27z\ge 54 \\ &\iff z\ge 6 \\ &\iff y^3-12y\ge 16 \\ &\iff y\ge 4 \\ &\iff x^3-3x\le 2 \\ &\iff x\le 2, \end{align*}so $x=2$. This easily gives that $(x,y,z)=(2,4,6)$, as desired.

We have \begin{align*} x\ge 2 &\implies \frac{z^3-27z}{27} \ge 2 \implies z^3-27z-54\ge 0 \implies z\ge 6 \text{ or } z=-3. \\ z\ge 6 &\implies y^3-12y = 3z-2 \ge 16 \implies y^3-12y-16 \ge 0 \implies y\ge 4 \text{ or } y=-2. \\ y\ge 4 &\implies x^3-3x-50=-12y \le -48 \implies x^3-3x-2\le 0 \implies x\le 2 \text{ or } x=-1. \end{align*}It is easy to check that none of $z=-3$ or $y=-2$ or $x=-1$ produce valid solutions. Therefore, $x\ge 2 \implies x\le 2$. We can make a similar inequality chain to show that $x\le 2\implies x\ge 2$. Therefore, $x=2$, which produces the solution $(x,y,z)=(2,4,6)$.

Substituting $x=a+2$, $y=b+4$, $z=c+6$, we get the equations $a(a+3)^2=-12b$, $b(b+6)^2=3c$, $c(c+9)^2=27a$. So, $a,b$ are of opposite sign, $b,c$ are of the same sign, and $c,a$ are of the same sign. The only way this can happen is when $a=b=c=0$, so our only solution is $(2,4,6)$.

hmm my solution seems to be identical to william122's but whatever I claim that $(x, y, z)=(2, 4, 6)$ are the only solutions. It is trivial to verify that these work. To prove that they are the only ones, let $x-2=a$, $y-4=b$, $z-6=c$. The equations rearrange to $a(a+3)^2=-12b, b(b+6)^2=3c$, and $c(c+9)^2=27a$. If one of $(a, b, c)$ is $0$ then clearly they are all $0$. If they are all nonzero, then note $\frac{a}{b}<0$, $\frac{b}{c}>0$, and $\frac{c}{a}>0$, but by looking at signs, this is clearly impossible. Hence, $(a, b, c)=(0, 0, 0)$, so we are done. $\blacksquare$

The answer is $(x, y, z) = (2, 4, 6)$, which we will show is the only valid solution. Set $x = a + 2$, $y = b+4$, and $z = c + 6$. Expanding the given equations gives $a(a+3)^2 = -12b, b(b+6)^2 = 3c$, and $c(c+9)^2 = 27a$. If $a = 0$, then the solution set follows. Else, if $a > 0$, then we must have $b < 0$, meaning $c < 0$, which in turn implies $a < 0$, contradiction. Otherwise, if $a < 0$, then $b > 0$, which means $c > 0$, so $a > 0$, another contradiction. $\blacksquare$

doing equality i see

The answer is $\boxed{(x,y,z) = (2,4,6).}$ Now let $a = x-2, b = y-4, z - x-6$. Substituting, the problem tells us that $a(a+3)^2 = -12b, b(b+6)^2 = 3c,$ and $c(c+9)^2 = 27a$. Note that clearly either one of $(a,b,c) =0$ forces all three to be $0$, from which we derive our equality set. Otherwise, note that $a$ and $b$ have opposite signs, $b$ and $c$ have the same sign, and $a$ and $c$ have the same sign, so $a$ and $b$ must have the same sign, a contradiction. Remarks: Hopefully not all equality problems end up like this

Great problem! Notice the equality case (2,4,6), hence we are motivated to let $x=a+2,y=b+4,z=c+6$ and try to show they're all 0. We get $$a^3+6a^2+12a+8=(3a+6)-(12b+48)+50\iff a^3+6a^2+9a=-12b\iff a(a+3)^2=-12b,$$$$b^3+12b^2+48b+64=(12b+48)+(3c+18)-2\iff b^3+12b^2+36b=3c\iff b(b+6)^2=3c,$$$$c^3+18c^2+108c+216=(27c+162)+(27a+54)\iff c^3+18c^2+81c=27a\iff c(c+9)^2=27a.$$Multiplying them together, the LHS has sign abc while the RHS has opposite sign -abc; hence abc=0. Now, caseworking on any one of them being 0, we get that each of a,b,c=0, as desired. $\blacksquare$

The only triplet is $(x, y, z) = (2, 4, 6)$, which works. To prove this is the only triplet, let $x = a + 2, y = b + 2, z = c + 2$. Then our equations rewrite as $$\begin{aligned} a(a + 3)^2 &= -12b, \\ b(b + 6)^2 &= 3c, \\ c(c + 9)^2 &= 27a. \end{aligned}$$If any of $a, b, c$ equals $0$, it's easy to see they must all equal $0$ (which works). Else, If $a > 0$, then $b < 0$ from the first equation, and $c < 0$ from the second equation, which violates the third equation If $a < 0$, then $b > 0$ from the first equation, and $c > 0$ from the second equation, which violates the third equation Thus, $(a, b, c) = (0, 0, 0) \implies (x, y, z) = (2, 4, 6)$, as desired.

We claim that the only triple is $(x,y,z) = \boxed{(2,4,6)}$, which clearly works. Substitute in $(x,y,z)=(a+2, b+4, c+6)$ to get \begin{align*} a(a+3)^2 &= -12b,\\ b(b+6)^2 &= 3c,\\ c(c+9)^2 &= 27a. \end{align*} If $a=0$, then we have $b=c=0$, which produces our claimed solution. If $a>0$, we have $b<0 \implies c<0 \implies a<0$, contradiction. If $a<0$, we have $b>0 \implies c>0 \implies a>0$, contradiction. Thus, there are no other solutions than the one claimed.

USA TST 2009/7 wrote: wrote: Find all triples $ (x,y,z)$ of real numbers that satisfy the system of equations \[ \begin{cases}x^3 = 3x-12y+50, \\ y^3 = 12y+3z-2, \\ z^3 = 27z + 27x. \end{cases}\] Fun(ny?) Problem realised never typed this up re-write the equations as follows $(x-2)(x+1)^2=-12(y-4)$ $(y-4)(y+2)^2=3(z-6)$ $(z-6)(z+3)^2=27(x-2)$ if $x \neq 2, y \neq 4, z \neq 6$ then multiply all of the equations, we get one side positive while other negative, contradiction!. Hence forcing $x=2,y=4,z=6$

We claim that the only solution is $\boxed{(x,y,z)=(2,4,6)}$. Let $x=a,y=2b,z=3c$. Substituting them, we get the following equations: \[ \begin{cases}a^3-3a=50-4b, \\ b^3-3b=\frac{9c-2}8, \\ c^3-3c=a. \end{cases}\]Note that the function $f(x)=x^3-3x$ is a cubic polynomial function. $f'(x)=0$ has two solutions, namely $x=\pm1$. $f(1)=-2,f(-1)=f(2)=2$. If $x<-1$ then $f(x)<f(-1)=f(2)=2$ and if $-1<x<1$, then $2=f(-1)>f(x)>f(1)=-2$. If $1<x<2$ then $f(x)<f(2)$. This means that if $x<2$ then $f(x)\leq f(2)$ and if $x>2$ then $f(x)>f(2)$. So, if $a>2$, then $50-24b>2$, giving $b<2$. If $b<2$, then $\frac{9c-2}8\leq2$, which means $c\leq2$. If $c\leq2$ then $a\leq2$, contradiction. If $a<2$, then $50-24b<2$, giving $b>2$. If $b>2$, then $\frac{9c-2}8>2$, which means $c>2$. If $c>2$ then $a>2$, contradiction. This means $a=2$ is forced, giving $b=2$ and $c=2$, which clearly work. So, $(x,y,z)=(2,4,6)$ is the only solution.

Answer is $(2, 4, 6)$. Substitute in \[(x, y, z) = (\alpha + 2, \beta + 4, \gamma + 6)\]\[(\alpha + 2)^3 = 3\alpha - 12\beta + 8\]\[(\beta + 4)^3 = 12\beta + 3\gamma + 64\]\[(\gamma + 6)^3 = 27\alpha + 27\gamma + 216\]\[\implies\]\[\alpha(\alpha +3)^2 = -12\beta\]\[\beta(\beta + 6)^2 = 3\gamma\]\[\gamma(\gamma + 9)^2 = 27\alpha\]\[\implies\]\[\alpha\beta\gamma(\alpha +3)^2(\beta + 6)^2(\gamma + 9)^2 = -972\alpha\beta\gamma\]FTSOC, assume that $\alpha\beta\gamma \neq 0$. Then dividing by $\alpha\beta\gamma$ results in \[(\alpha +3)^2(\beta + 6)^2(\gamma + 9)^2\ = -972\]which is a contradiction due to the trivial inequality. Hence, $\alpha\beta\gamma = 0$. Plugging $0$ into $\alpha$, $\beta$, or $\gamma$, we can clearly see that it makes the rest of the variables $0$, so our answer for $(x, y, z)$ is $(2, 4, 6)$.

We claim our only solution is $\boxed{(x,y,z)=(2,4,6)}$. Motivated by this answer, we notice our equations can be rearranged to be \begin{align*} (x-2) \cdot ((x-2)+5)^2 = -12(y-4) \\ (y-4) \cdot ((y-4)+10)^2 = 3(z-6) \\ (z-6) \cdot ((z-6)+15)^2 = 27(x-2). \end{align*} Multiplying these, we see that we must have one of $x=2$, $y=4$, or $z=6$ to avoid a trivial inequality contradiction. However, any one of the these imply the other two, which finishes. $\blacksquare$

We claim that $(x, y, z) = (2, 4, 6)$ which clearly works. Note the system of equations rearranges to \begin{align*} (x + 1)^2 (x - 2) &= -12(y - 4) \\ (y + 2)^2 (y - 4) &= 3(z - 6) \\ (z + 3)^2 (z - 6) &= 27(x - 2). \end{align*}Multiplying yields \[(x + 1)^2 (y + 2)^2 (z + 3)^2 (x - 2)(y - 4)(z - 6) = -972(x - 2)(y - 4)(z - 6).\]Now if $x = 2$, $y = 4$ or $z = 6$ then it is easily seen that $(x, y, z) = (2, 4, 6)$ by direct substitution. Otherwise \[(x + 1)^2 (y + 2)^2 (z + 3)^2 = -972,\]a contradiction.

The only solution is $\boxed{(2, 4, 6)}$. Consider rearranging as, \begin{align*} (x-2)(x+1)^2 &= 12(4 - y)\\ (y-4)(y+2)^2 &= 3(z - 6)\\ (z-6)(z+3)^2 &= 27(x - 2) \end{align*}Multiplying we have, \begin{align*} (x-2)(y-4)(z-6)(x+1)^2(y+2)^2(z+3)^2 &= -927(x-2)(y-4)(z-6) \end{align*}Clearly if all of the factors $(x-2)$, $(y-4)$ and $(z-6)$ are nonzero we have no solutions. However if we have $x = 2$, $y = 4$ or $z = 6$ we find the claimed solution so we are done.

The hardest part of this problem is finding the solution itself. We claim that the only solution to this system of equations is $(x,y,z) = (2,4,6).$ This can clearly be shown to work just by plugging in. Now we show that this is the only solution. Inspired by the equality case, we set $x = p + 2, y = q + 4,$ and $z = r+6.$ After plugging in and reducing the equation, we get the following systems of equations: \begin{align*} p(p+3)^2 &= -12q \\ q(q+6)^2 &= 3r \\ r(r+9)^2 &= 27p. \end{align*}If any of $p,q,r$ are equal to $0,$ then all of them must be $0$ simply by plugging in, giving us the solution claimed at the beginning of our solution. Henceforth assume that $p,q,r$ are nonnegative. The first equation rearranges to \[ -\frac{p}{12q} = \left(\frac{1}{p+3}\right)^2, \]so \[ -\frac{81p}{q} = 972 \cdot \left(\frac{1}{p+3}\right)^2. \]Meanwhile, the second and third equations multiply to \[ qr((q+6)(r+9))^2 = 81pr, \]which, since $r \ne 0,$ rearranges to \[ \frac{81p}{q} = ((q+6)(r+9))^2. \]Adding our two equations together, we get \[ 972 \cdot \left(\frac{1}{p+3}\right)^2 + ((q+6)(r+9))^2 = 0. \]Since the square of a real number is nonnegative, all of the squares in the above expression are $0.$ In particular, $\frac{1}{p+3} = 0,$ which is impossible. Therefore, $p = q = r = 0,$ and so the only solution to our system of equations is $\boxed{(x,y,z) = (2,4,6)},$ as claimed.

Answer: $(2,4,6).$ We can check this works. Substitute $x=X+2,y=Y+4,z=Z+6.$ The equations become \begin{align*}&X(X+3)^2=-12Y,\\&Y(Y+6)^2=3Z,\\&Z(Z+9)^2=27X.\end{align*}Now it is easy to see that if one of $X,Y,Z$ is zero then all are. But multiplying all together gives $-972XYZ=XYZ(X+3)^2(Y+6)^2(Z+9)^2,$ impossible if $XYZ\ne 0.$ Thus $(X,Y,Z)=(0,0,0)$ so $(x,y,z)=(2,4,6).$

The solution $(x, y, z) = (2, 4, 6)$ is the only one. Now note that the equations are equivalent to: $$(x + 1)^2 (x - 2) = -12(y - 4)$$$$(y + 2)^2 (y - 4) = 3(z - 6)$$$$(z + 3)^2 (z - 6) = 27(x - 2)$$It is easy to see that $x = 2$, $y = 4$, or $z = 6$ are independently sufficient enough to yield $(2, 4, 6)$ as a solution, so assume $x \ne 2, y \ne 4, z \ne 6$. Then multiplying the three equations and dividing by $(x - 2)(y - 4)(z - 6)$ gives $$(x + 1)^2 (y + 2)^2 (z + 3)^2 = -12 \cdot 3 \cdot 27$$which is clearly impossible. So $(x, y, z) = (2, 4, 6)$ is the only solution and we are finished.

We claim the only solution is $(x, y, z) = (2, 4, 6)$, which clearly works. Let $x = a+2, y = b+4, z = c+6$. Then the equations simplify as $$a(a+1)^2 = -12b$$$$b(b+2)^2 = 3c$$$$c(c+3)^2 = 27a.$$Now assume none of $a, b, c$ are zero, since if one is then clearly all are. Then we get $$(a+1)^2 = -\frac{12b}{a}, (b+2)^2 = \frac{3c}b, (c+3)^2 = \frac{27a}c.$$In particular, if $sgn(k)$ denotes the sign of $k$, $$sgn(a) \ne sgn(b), sgn(b) = sgn(c), sgn(c) = sgn(a),$$a clear contradiction. Therefore the only working solution is $(x, y, z) = (a+2, b+4, c+6) = (2, 4, 6)$. $\square$

Answer: $(x,y,z)=(2,4,6)$ Solution: Let $x=a+2$ , $y=b+4$ , $z=c+6$ Our system of equations transforms into the following: \[ \begin{cases}a^3+6a^2+9a=-12b ...(1) , \\ b^3+12b^2+36b=3c ...(2) , \\ c^3+18c^2+81c=27a ...(3) \end{cases}\] Since we want to show that $a=b=c=0$ we will prove it by a contradiction. AFTSOC $a \neq 0$ $\textbf{Case 1.}$ $a >0$ $...(*)$ From $(1) \implies -12b \stackrel{(1)}{=}a^3+6a^2+9a >0+0+0=0 \implies -12b>0 \implies -b>0 \implies b<0$ $...(**)$ Combining $(2)$ and $(**) \implies 0=0+0+0 \stackrel{(**)}{<} b^3+12b^2+36b \stackrel{(2)}{=} 3c \implies 0<3c \implies 0<c$ $...(***)$ Combining $(3) , (*) $ and $(***)$ we get: $0=0+0+0 \stackrel{(***)}{<} c^3+18c^2+81c \stackrel{(3)}{=} 27a \stackrel{(*)}{<} 27 \cdot 0 =0 \implies 0<0 \rightarrow \leftarrow $. $\textbf{Case 2.}$ $a<0$ $...(@)$ From $(1) \implies -12b \stackrel{(1)}{=}a^3+6a^2+9a < 0+0+0=0 \implies -12b<0 \implies -b<0 \implies b>0$ $...(@@)$ Combining $(2)$ and $(@@) \implies 0=0+0+0 \stackrel{(@@)}{>} b^3+12b^2+36b \stackrel{(2)}{=} 3c \implies 0>3c \implies 0>c$ $...(@@@)$ Combining $(3) , (@)$ and $(@@@)$ we get: $0=0+0+0 \stackrel{(@@@)}{>} c^3+18c^2+81c \stackrel{(3)}{=} 27a \stackrel{(@)}{>} 27 \cdot 0 =0 \implies 0>0 \rightarrow \leftarrow $. Hence $\boxed{a=0}$ from $(1)$ we find that $\boxed{b=0}$ and finally from $(2)$ we find that $\boxed{c=0}$. $\blacksquare$