We begin by manipulating the given equations to find other useful relations between the variables. $ (\sum x)^2 = \sum x^2 + 2 \sum xy$ $ 3^2 = 3 + 2 \sum xy$ $ \sum xy = 3$ Now using the following identity, $ \sum x^3 - 3xyz = (\sum x)(\sum x^2 - \sum xy)$ $ 3 - 3xyz = 3(0)$ $ xyz = 1$ Now, let $ x,y,$ and $ z$ be the roots of a cubic polynomial $ P(t)$. Then, by Vieta's Formulas, we have that $ P(t) = t^3 - 3t^2 + 3t - 1$ $ P(t) = (t - 1)^3$ Therefore, $ \boxed{x = y = z = 1}$ is the only solution.

If we square the $ x + y + z = 3$, then we get $ 9 = x^2 + y^2 + z^2 + 2(xy + yz + xz)$ $ 9 = 3 + 2(xy + yz + xz)$ $ xy + yz + xz = 3$ If we multiply this by $ x + y + z$, then we have $ 9 = (x^2y + xy^2 + y^2 + yz^2 + x^2z + xz^2) + 3xyz$ and we substitute the contents in the parenthesis with $ a$ for convenience, then we have $ 9 = a + 3xyz$ $ xyz = \frac {9 - a}{3}$ If we multiply $ x^2 + y^2 + z^2$ by $ x + y + z$, then we get $ x^3 + y^3 + z^3 + (x^2y + xy^2 + y^2 + yz^2 + x^2z + xz^2) = 9$ $ x^3 + y^3 + z^3 + a = 9$ $ 3 + a = 9$ $ a = 6$ If we plug this back in to get xyz, we have $ xyz = 1$ Notice that we now have $ x + y + z$, $ xy+yz+xz$, and $ xyz$ and if we use $ x$, $ y$, and $ z$ as roots of a cubic equation, then we have $ a^3 + 3a^2 + 3a + 1 = 0$ $ (a - 1)^3 = (a - x)(a - y)(a - z)$ Therefore $ x = y = z = 1$ and I guess that's the only solution... correct me if I'm wrong please... oh yeah, modularmarc101 got the same thing so I guess I can't be that off...

daydreamer1116 wrote: Notice that we now have $ x + y + z$, $ x^2y + xy^2 + y^2 + yz^2 + x^2z + xz^2$, and $ xyz$ and if we use $ x$, $ y$, and $ z$ as roots of a cubic equation, then we have... I think you meant $ xy+yz+zx$ for the middle one. If so, then the solution is correct.

Thanks. I think I'm good now

Well, everyone else got $x=y=z=1$, and I agree, so I'm not sure what you mean. Computer confirms that that is the only solution.

professordad wrote: We have $\left(\sum x\right)^2 = \left(\sum x^2\right) + 2\left(\sum xy\right)$, so $xy + yz + xz = 3$. Now $x^2 + y^2 + z^2 - xy - yz - xz = 0$, so $(x - y)^2 + (y - z)^2 + (x - z)^2 = 0$. All squares must equal 0, so $x = y = z = 1$. When $x,y$ and $z$ are complex numbers, we don't immediately have \[(x-y)^2\ge 0\] (and cyclic permutations) So it's not immediate that \[(x - y)^2 + (y - z)^2 + (x - z)^2 = 0\implies x=y=z\]

You can use the Chebyshev's ineguality and $\underline{(x+y+z)(x^2+y^2+z^2)}=3\cdot 3=\underline{3\cdot (x^3+y^3+z^3)}\iff x=y=z$ .

You cannot use Chebyshev's inequality (which makes no sense) when the variables might be complex numbers.

OK. I don't see complex numbers, sorry ... Thank you. Happy new year !

Let $P(x)=t^3+at^2+bt+c=0$ be a cubic polynomial with roots $x,y,z.$ By Vieta's we know that $$a=-1(x+y+z) \implies a=-3.$$We also know that $b=xy+xz+yz.$ We can now use Newton's sums to derive $b.$ Newton's sums tells us that $$(x+y+z)(x+y+z)-2(xy+xz+yz)=x^2+y^2+z^2.$$We can plug in the values of $x+y+z$ and $x^2+y^2+z^2$ to get $$3\cdot 3-2(xy+xz+yz)=3 \implies xy+xz+yz=3.$$Hence, $b=3.$ We can use Vieta's and Newton's sums again to derive $c$ as we know that $c=-1(xyz).$ So, by Newton's sums, it follows that $$(x+y+z)(x^2+y^2+z^2)-(xy+xz+yz)(x+y+z)+3(xyz)=x^3+y^3+z^3.$$We know what $(x+y+z), (x^2+y^2+z^2), (xy+xz+yz)$ and $(x^3+y^3+z^3)$ are so we plug them in $$3\cdot 3-3\cdot 3+3(xyz)=3 \implies xyz=1.$$Hence, $c=-1.$ Now our cubic polynomial is $t^3-3t^2+3t-1=0.$ That factors into $(t-1)^3=0.$ So, $t=1$. Therefore, our only roots to the system of simultaneous equations are $\boxed{x=y=z=1}.$

can you use the newton sum thing?

Louis_Vuitton wrote: can you use the newton sum thing? Yes! The answer is $(x,y,z)=(1,1,1)$ which clearly works. Now we will show that it is the only solution. Let $x,y,z$ be the roots of some cubic polynomial $x^3+ax^2+bx+c$. Define the sums $P_k=x^k+y^k+z^k$. Then by Newton Sums, \begin{align*}P_1+a&=0\\P_2+aP_1+2b&=0\\ P_3+aP_2+bP_1+3c&=0\end{align*}or \begin{align*}a&=-P_1\\b&=-\frac{P_2+aP_1}2\\ c&=-\frac{P_3+aP_2+bP_1}3.\end{align*} We are already given $P_1=P_2=P_3=3$, so we can find $a=-3$ which tells us that $b=3$ which tells us that $c=-1$. Hence, $x,y,z$ are the solutions to $$f(t)=t^3-3t^2+3t-1.$$Factoring, $$f(t)=t^3-3t^2+3t-1=(t-1)^3$$so $f(x)$ has a triple root at $t=1$. Thus, we find that $(x,y,z)=(1,1,1)$ is the only solution to the original system.

We claim that the only root is $x=y=z=1$, which clearly works.Note that \[xy+yz+xz = \frac12 \left((x+y+z)^2-x^2-y^2-z^2\right) = 3\] \[3xyz = (x^3+y^3+z^3) - (x+y+z)(x^2+y^2+z^2-xy-yz-xz) = 3 - 3\cdot 0 =3\] Now, note that $x,y,z$ are the roots of the polynomials \[w^3-3w^2+3w-1 = (w-1)^3\]Thus, we must have $(x,y,z)=1$ and we're done. $\blacksquare$

USAMO was much easier in 1973 this only took me a few minutes to solve (though the idea was similar to DAMO P6) Let $a=xy+yz+xz$ and $b=xyz$. Since $x+y+z=3$, $x,y,z$ are the roots of the polynomial $X^3-3X^2+aX-b$. We have $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=9-2a$, so $9-2a=3\implies a=3$. Now, $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0\implies 3-3b=0\implies b=1$. Thus, $x,y,z$ are roots of the equation $X^3-3X^2+3X-1=(X-1)^3$, so $\boxed{(x,y,z)=(1,1,1)}$, which clearly works.

Using Newton's Formulas notation, we have that \begin{align*} \rho_1&=\sigma_1=3 \\ \rho_2&=\sigma_1^2-2\sigma_2=3 \\ \rho_3&=\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3 \\ \end{align*}Routinely solving each equation one at a time, we get that $\sigma_1=3, \sigma_2=3, \sigma_3=1$. By Vieta's Formulas, we know that $x$, $y$, and $z$ are the roots of the polynomial $x^3-3x^2+3x-1=0$. This can be factored as $(x-1)^3=0$, so clearly $(x,y,z)=(1,1,1)$ is the only solution which we can check works.

Nowadays this would be suitable for AIME. Define $P_k$ the $k$th power sum polynomial and similarly define $S_k$ the $k$th symmetric sum polynomial. By Newton-Girard Identities, we have the relations \begin{align*} P_1-S_1=0, \\ P_2-S_1P_1+2S_2=0, \\ P_3-S_1P_2+S_2P_1-3S_3=0. \end{align*}Solving we see $S_1=x+y+z=3, S_2=xy+yz+zx=3, S_3=xyz=1.$ Then, we can construct the polynomial $p(t)$ with roots $x,y,z$ as $p(t)=t^3-3t^2+3t-1,$ and it follows the roots are $(x,y,z)=(1,1,1),$ the desired triple.

Brut3Forc3 wrote: Determine all roots, real or complex, of the system of simultaneous equations \begin{align*} x+y+z &= 3, \\ x^2+y^2+z^2 &= 3, \\ x^3+y^3+z^3 &= 3.\end{align*} A classic

Denote $a, b,$ and $c$ as follows, \begin{align*} a & = x+y+z, \\ b & = xy+yz+zx, \\ c & = xyz. \end{align*}Our given equations become \begin{align*} 3 & = a, \\ 3 & = a^2-2b, \\ 3 & = a(a^2-b)+3c =a^3-ab+3c. \end{align*}Solving for $b$ and $c$, we obtain $b=3$ and $c=1$. Now let $P(x)$ be the cubic polynomial with roots $x, y,$ and $z$. By Vieta's, \begin{align*} P(x) & =x^3-3x^2+3x-1, \\ & =(x-1)^3. \end{align*}Therefore, $\boxed{x=y=z=1}$ is our only solution.

Let $P(t)=t^3+a_2t^2+a_1t+a_0$ the monic pollynomial with roots $x,y,z$; by Newton Sums \begin{align} 0 &= 1 \cdot (x+y+z)+a_2\Rightarrow a_2=-3\nonumber \\ &=1\cdot (x^2+y^2+z^2)+a_2(x+y+z)+a_1\cdot 2\Rightarrow a_1=3 \nonumber \\ &=1\cdot(x^3+y^3+z^3)+a_2(x^2+y^2+z^2)+a_1(x+y+z)+3a_0\Rightarrow a_0=-1\nonumber \end{align}Therefore $P(t)=t^3-3t^2+3t-1=(t-1)^3\Rightarrow x=y=z=1$

Let, $q=x+y+z, r=xy+xz+yz, s=xyz,$ then $3=q, 3=q^2-2r, 3=q^3-qr+3s.$ Note, that $r=\frac{q^2-3}{2}=\frac{9-3}{2}=3.$ And, we find that $s=1.$ Now, by Vieta's this is the cubic polynomial, $x^3-3x^2+3x-1=(x-1)^3$, so we must have $x=y=z=1.$

We will use standard SQP notation: \begin{align*} s &= x+y+z \\ q &= xy+yz+zx \\ p &= xyz \end{align*} From the first equation, we immediately know $s=3$. Then, we have $x^2+y^2+z^2=s^2-2q$, so we can solve that $q=3$. To apply the last equation, notice that \[x^3+y^3+z^3=3xyz+(x+y+z)(x^2+y^2+z^2-xy-yz-xz)\] Simplifying this and solving, we get $p=1$. By Vieta's, we see that $x$, $y$, and $z$ are the roots of polynomial $P(t)=t^3-3t^2+3t-1=(t-1)^3$. Thus, the only triple $(x,y,z)$ is $\boxed{(1,1,1)}$.

By using Newton's sum, we can find that $x$,$y$ and $z$ are roots of the polynomial $P(t)=(t-1)^3$. This implies that $x=y=z=1$ is the only solution to this system of equations.

I claim that the only possible solution is $(1,1,1)$. Note that from the second equation, we get that \[xy+yz+zx=\frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\frac{9-3}{2}=3,\]and using this, along with the third equation, we get that \[3-3xyz=(x^3+y^3+z^3)-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=3(0)=0 \iff xyz=1,\]which gives us the system of equations \[x+y+z=3,\]\[xy+yz+zx=3,\]\[xyz=1,\]which implies that $x$, $y$, $z$ should all be roots of $f(a)=a^3-3a^2+3a-1=(a-1)^3$ by Vieta's. This gives us that the only solution to $(x,y,z)$ is $(1,1,1)$, finishing the problem.

since, $x+y+z=3, x²+y²+z²=3$, we can say $xy+yz+zx = 3$. then, by subtracting $3xyz$ by both sides by the third equation, $$x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)=0$$also, $x³+y³+z³=3$ so $xyz=1$. therefore we know all of the sum, pair sum, product of $x, y, x$, we can know these 3 variables are the roots of a polynomial $t³-3t²+3t-1=0$ therefore the answer is $\boxed{x=y=z=1}$

From equation $(1)$ and equation $(2)$, we obtain $xy+yz+zx=3$. It follows that $x^2+y^2+z^2=xy+yz+zx$, and multiplying both sides by $2$ and rearranging to factor into binomials yields$$(x-y)^2+(y-z)^2+(z-x)^2 = 0$$so we must have $x=y=z=1$ as the only solution.

No way is this USAMO. From standard identities for $(a+b+c)^2$ and $a^3 + b^3 + c^3 -3abc$, we can obtain $$xy + yz + zx = 3, xyz = 1,$$so $x, y, z$ are the zeroes of $$t^3 - 3t^2 + 3t - 1 = (t-1)^3,$$and we get $$x = y = z = 1.$$Since it works, we're done.$\square$

blueberryfaygo_55 wrote: From equation $(1)$ and equation $(2)$, we obtain $xy+yz+zx=3$. It follows that $x^2+y^2+z^2=xy+yz+zx$, and multiplying both sides by $2$ and rearranging to factor into binomials yields$$(x-y)^2+(y-z)^2+(z-x)^2 = 0$$so we must have $x=y=z=1$ as the only solution. How do you get the last step when $x, y, z$ need not be real?

Lets assume that x, y, z are the roots of a cubic polynomial $ax^3 + bx^2 + cx + d$. By the condition we have $p_1 = 3$, $p_2 = 3$, $p_3 = 3$ $\Rightarrow$ by Newton's sums we have that 0 = 3a + b = 3a + 3b + 2c = 3a + 3b + 3c + 3d. Also we can assume a = 1 $\Rightarrow$ from this we get b = -3, c = 3, d = -1 $\Rightarrow$ the polynomial which roots are x, y and z is $x^3 - 3x^2 + 3x - 1$. Also $x^3 - 3x^2 + 3x - 1 = (x - 1)^3$ $\Rightarrow$ x = y = z = 1.

Consider the polynomial $(t-x)(t-y)(t-z) = 0$. The $t^2$ coefficient is $-3$ here by vietas and the $t$ coefficient is $\frac{3^2-3}{2} = 3$ by vietas as well. Also $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$ so we can find that $xyz=1$. Thus the polynomial is $x^3-3x^2+3x-1 = (x-1)^3$ so $\boxed{x=y=z=1}$ is the only solution.

The answer is $(1,1,1)$ only. Let $$a=x+y+z$$$$b=xy+yz+zx$$$$c=xyz.$$Note that $a=3$, $a^2-2b=3$, and $$3-3c=a(a^2-3b)\implies (a,b,c)=(3,3,1).$$By Vieta's, we find that $x$, $y$, and $z$ are the roots of the cubic $$a^3-3a^2+3a-1=(a-1)^3=0,$$so $(x,y,z)=(1,1,1)$ is the only solution.