If$$x+y+z=1$$ $$x^2+y^2+z^2=2$$ $$x^3+y^3+z^3=3$$ Then find the value of $$x^5+y^5+z^5$$
Is there any simple way to solve this problem ? I have tried all my tricks tried to multiply two equations , substitute $z=1-x-y$ , but things got messy nothing seems to work out .
This form of equations can be solved systematically using Newton's identities.
The problem at hand is a special case of $3$ variables. Let $e_1,e_2,e_3$ be the three elementary symmetric polynomial associated with $x,y,z$: $$\begin{align} e_1 &= x + y + z\\ e_2 &= xy+ yz + zx\\ e_3 &= xyz \end{align}$$ This is equivalent to $\displaystyle\;(\lambda - x)(\lambda -y)(\lambda -z) = \lambda^3 - e_1\lambda^2 + e_2\lambda - e_3$.
For any integer $k > 0$, let $p_k = x^k + y^k + z^k$. For 3 variables, the Newton's identities are:
$$\begin{align} p_1 &= e_1\\ p_2 &= e_1 p_1 - 2 e_2\\ p_3 &= e_1 p_2 - e_2 p_1 + 3e_3\\ p_n &= e_1 p_{n-1} - e_2 p_{n-2} + e_3 p_{n-3} \quad\text{ for }\; n > 3 \end{align}$$
Given the known values of $(p_1,p_2,p_3) = (1,2,3)$, we have
$$\begin{align} e_1 &= p_1 = 1;\\ e_2 &= -\frac12(p_2 - e_1p_1) = -\frac12\left(2 - 1^2\right) = -\frac12\\ e_3 &= \frac13(p_3 - e_1p_2 + e_2p_1) = \frac13\left(3 - 2 - \frac12\right) = \frac16 \end{align}$$
Substitute this into formula of $p_4$ and $p_5$, we have
$$\begin{align} p_4 &= e_1 p_3 - e_2 p_2 + e_3 p_1 = 3 + \frac12\cdot 2 + \frac16\cdot 1 = \frac{25}{6}\\ p_5 &= e_1 p_4 - e_2 p_3 + e_3 p_2 = \frac{25}{6} + \frac12\cdot 3 + \frac16\cdot 2 = 6 \end{align}$$
The answer is $x^5 + y^5 + z^5 = 6$.
HINT
Multiply the third relation by the second so that you obtain \begin{align*} (x^{2} + y^{2} + z^{2})(x^{3} + y^{3} + z^{3}) & = x^{5} + y^{5} + z^{5} + x^{2}(y^{3} + z^{3}) + y^{2}(x^{3} + z^{3}) + z^{2}(x^{3} + y^{3}) \end{align*}
Now we can rearrange the last expression as \begin{align*} x^{2}y^{2}(x + y) + x^{2}z^{2}(x + z) + y^{2}z^{2}(y + z) & = x^{2}y^{2} + x^{2}z^{2} + y^{2}z^{2} - x^{2}y^{2}z - x^{2}yz^{2} - xy^{2}z^{2}\\\\ & = x^{2}y^{2} + x^{2}z^{2} + y^{2}z^{2} - xyz(xy + xz + yz) \end{align*}
Hence the problem has been reduced to study the last relation.
Before keep going, notice that \begin{align*} x^{2}y^{2} + x^{2}z^{2} + y^{2}z^{2} & = (xy + xz + yz)^{2} - 2xyz(x + y + z)\\\\ & = (xy + xz + yz)^{2} -2xyz \end{align*}
In order to find the remaining expression, let us observe that \begin{align*} (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2(xy + xz + yz) \Rightarrow xy + xz + yz = -\frac{1}{2} \end{align*}
At last but not least, we have to find the value of $xyz$.
Can you take it from here?
Here is yet another way, in case you are interested: $$(x+y+z)^2=1\implies\sum x^2+2\sum xy=1\implies \sum xy=-\frac12$$ Therefore the polynomial equation whose roots are $x,y,z$ takes the form: $$t^3-t^2-\frac12t+c=0$$ Summing this equation for each of the roots, $$\sum x^3-\sum x^2-\frac12\sum x +3c=0$$ $$\implies3-2-\frac12(1)+3c=0\implies c =-\frac16$$ Multiplying the polynomial by $t$ and summing again, $$\sum x^4-\sum x^3-\frac12\sum x^2-\frac16\sum x=0$$ $$\implies\sum x^4=3+\frac12(2)+\frac16(1)$$ $$\implies \sum x^4=\frac{25}{6}$$ Multiplying the polynomial by $t^2$ and summing again, $$\sum x^5-\sum x^4-\frac12\sum x^3-\frac16\sum x^2=0$$ $$\implies\sum x^5=\frac{25}{6}+\frac12(3)+\frac16(2)=6$$
Another way.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$,where $v^2$ can be negative, and $xyz=w^3$.
Thus, $u=\frac{1}{3}$ and since $$2=x^2+y^2+z^2=9u^2-6v^2,$$ we obtain $$v^2=-\frac{1}{6}.$$ Also, since $$3=x^3+y^3+z^3=27u^3-27uv^2+3w^3=1+\frac{3}{2}+3w^3,$$ we get $$w^3=\frac{1}{6}.$$ Id est, $$x^5+y^5+z^5=243u^5-405u^3v^2+135uv^4+45u^2w^3-15v^2w^3=6.$$ I used the following identities, which easy to prove and which we always use in the $uvw$'s method.
About $uvw$ see here https://artofproblemsolving.com/community/c6h278791
$$x^2+y^2+z^2=9u^2-6v^2,$$ $$x^3+y^3+z^3=27u^3-27uv^2+3w^3$$ and $$x^5+y^5+z^5=243u^5-405u^3v^2+135uv^4+45u^2w^3-15v^2w^3.$$ There are also: $$\sum_{cyc}(x^2y+x^2z)=9uv^2-3w^3,$$ $$\sum_{cyc}x^2y^2=9v^4-6uw^3$$ and more
