An algebra problem by Dev Sharma

$Since\quad x+y+z=0\\ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }=3xyz\\ xyz=6\\ Let\quad xy+yz+xz=a\\ Then\quad { (x+y+z) }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }+2a\\ or\quad P_{ 2 }=-2a\quad ..........(1)\\ Let\quad x,y,z\quad be\quad the\quad roots\quad of\quad equation\quad \\ { x }^{ 3 }+ax-6=0\quad \quad \quad \quad \{ \because \quad x+y+z=0\quad so\quad no\quad { x }^{ 2 }\quad term\} \\ { x }^{ 3 }=6-ax\\ Squaring\\ { x }^{ 6 }=36+{ (ax) }^{ 2 }-12ax\\ multiplying\quad by\quad x\\ { x }^{ 7 }=36x+{ a }^{ 2 }{ x }^{ 3 }-12a{ x }^{ 2 }\\ \sum { { x }^{ 7 }=36\sum { x } +{ a }^{ 2 }\sum { { x }^{ 3 }-12a\sum { { x }^{ 2 } } } } \\ 2058=0+18{ a }^{ 2 }+24{ a }^{ 2 }\quad \quad \quad \quad \quad \{ by\quad (1)\} \\ { a }^{ 2 }=49\\ { P }_{ 2 }>0\quad \therefore \quad a<0\\ \therefore \quad a=-7\\ { P }_{ 2 }=14\\ So\quad the\quad equation\quad becomes\quad \\ { x }^{ 3 }-7x-6=0\\ { x }^{ 3 }=7x+6\\ Cubing\\ { x }^{ 9 }=216+343{ x }^{ 3 }+126x(7x+6)\\ multiplying\quad by\quad x\\ { x }^{ 10 }=216x+343{ x }^{ 4 }+126*7*{ x }^{ 3 }+126*6*{ x }^{ 2 }\quad \\ \sum { { x }^{ 10 } } =216\sum { x } +(126*7)\sum { { x }^{ 3 } } +(126*6)\sum { { x }^{ 2 } } +343\sum { { x }^{ 4 } } \\ \quad \quad \quad \quad \quad \quad =0+(126*7*18)+(126*6*14)+343\sum { { x }^{ 4 } } \\ By\quad newton's\quad identity:\\ { P }_{ 4 }={ S }_{ 1 }{ P }_{ 3 }-{ S }_{ 2 }{ P }_{ 2 }+{ S }_{ 3 }{ P }_{ 1 }\\ \quad \quad =0+(7*14)+0\\ \quad \quad =98\\ So,\\ \sum { { x }^{ 10 } } =(126*7*18)+(126*6*14)+(343*14)\\ On\quad solving\quad \\ { x }^{ 10 }+{ y }^{ 10 }+{ z }^{ 10 }=\boxed { 60074 } .$

Synopsis : By Newton's Sum, we first find the value of $x^2 + y^2 + z^2$ . Then finish it off by considering the expression $(x^3 + y^3 + z^3)(x^7 + y^7 + z^7 )$ .

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Let $P_n$ denote the value of $x^n + y^n + z^n$ for positive integer $n$ , then $P_1 = 0, P_3 = 18, P_7 = 2058$ and we want to evaluate $P_{10}$ .

And let $S_n$ denote the symmetric sum of $x,y$ and $z$ . That is, $S_1 =x+y+z = P_1 = 0$ , $S_2 = xy + xz + yz$ , $S_3 = xyz$ .

Consider the equation $X^3 - S_1 X^2 + S_2 X - S_3 = 0$ . By Newton's sum,

$\begin{aligned} &&P_2 - \cancel{S_1 P_1} +2S_2 = 0 \Rightarrow \color{#20A900}{S_2 = -\dfrac12 P_2 } \\ && P_3 - \cancel{S_1 P_2} + \cancel{S_2 P_1} - 3S_3 = 0\Rightarrow \color{#20A900}{S_3 = 6} \\ && P_4 - \cancel{S_1 P_3} + S_2 P_2 -\cancel{ S_3 P_1} = 0 \Rightarrow P_4 = -S_2 P_2 \Rightarrow P_4 = \color{#20A900}{\dfrac12 P_2^2 } \\ && P_5 - \cancel{S_1 P_4} + S_2 P_3 - S_3 P_2 = 0\Rightarrow P_5 = -18S_2 + 6P_2 \Rightarrow \color{#20A900}{P_5 = 15P_2 } \\ && P_7 - \cancel{S_1 P_6} + S_2 P_5 - S_3 P_4 = 0 \Rightarrow P_7 = -S_2 (15P_2) + 6\left(\dfrac12 P_2^2\right) \Rightarrow \color{#20A900}{P_7 = \dfrac{21}2 P_2^2 } \end{aligned}$

Because $P_7 = 2058$ , solving for $P_2 > 0$ yields $P_2 = 14$ .

Now consider the expression $(x^3 + y^3 + z^3)(x^7 + y^7 + z^7)$ :

$\begin{aligned} P_3 \times P_7 &=& P_{10} + (xy)^3 (x^4 + y^4) + (xz)^3 (x^4 + z^4 ) + (yz)^3 (y^4 + z^4) \\ 18\times2058 &=& P_{10} + (xy)^3 (P_4 - z^4 ) + (xz)^3 (P_4 - y^4) + (yz)^3 (P_4 - x^4) \\ 37044 - P_{10} &=& P_4 \left [ (xy)^3 + (xz)^3 + (yz)^3 \right] - \cancel{(xyz)^3(x+y+z)} \\ &=&P_4 \left [ (xy+xz+yz)^3 - \cancel{3(x+y+z)(xy + xz + yz)} + 3(xyz)^2 \right ] \\ &=&P_4 \left [ S_2 ^3 +3 \cdot 6^2 \right ] \\ &=&-\frac1{16} P_2^2 \left( P_2^3 - 864 \right) \\ P_{10} &=& \boxed{60074} \\ \end{aligned}$

$x + y + z = 0 \Rightarrow x^3 + y^3 + z^3 = 3xyz \Rightarrow \boxed{xyz = 6}$

$\boxed{x^2 + y^2 + z^2 = k , xy + yz + zx = \frac{-k}{2}}$

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$(x + y + z)(x^3 + y^3 + z^3) = 0 = x^4 + y^4 + z^4 + xy(x^2 + y^2) + yz(y^2 + z^2) + zx(z^2 + x^2)$

$\Rightarrow x^4 + y^4 + z^4 +xy(k - z^2) + yz(k - x^2) + zx(k - y^2)$

$\Rightarrow x^4 + y^4 + z^4 - xyz(x + y + z) + k(xy + yz + zx)$

$\Rightarrow x^4 + y^4 + z^4 + (\frac{-k^2}{2}) = 0$

$\Rightarrow \boxed{x^4 + y^4 + z^4 = \frac{k^2}{2}}$

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$(xy + yz + zx)^2 = x^{2}y^{2} + x^{2}z^{2} + z^{2}y^{2} + 2xyz(x + y + z) = \frac{k^2}{4}$

$\Rightarrow \boxed{x^{2}y^{2} + x^{2}z^{2} + z^{2}y^{2} = \frac{k^2}{4}}$

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$(x^3 + y^3 + z^3)(x^4 + y^4 + z^4) = x^7 + y^7 + z^7 + x^{3}y^{3}(x + y) + x^{3}z^{3}(x + z) + z^{3}y^{3}(y + z)$

$\Rightarrow x^7 + y^7 + z^7 - xyz(x^{2}y^{2} + x^{2}z^{2} + z^{2}y^{2}) = 9k^2$

$\Rightarrow 2058 - (6)(\frac{k^2}{4}) = 9k^2$

Simplifying gives $\boxed{k = 14 , -14}$ . But we reject $k = -14$ as per the condition that $\boxed{x^2 + y^2 + z^2 > 0}$

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After this , we can form the cubic equation with roots $x , y , z$ as we know the elementary symmetric sums :

$x + y + z = 0 , xy + yz + zx = - 7 , xyz = 6$

$\Rightarrow a^3 - 7a - 6 = 0 \Rightarrow a = -1 , -2 , 3$

$\therefore (x , y , z) = (-1 , -2 , 3)$

$\therefore x^{10} + y^{10} + z^{10} = (-1)^{10} + (-2)^{10} + 3^{10} = \boxed{60074}$

This in not an actual solution.

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As $x+y+z=0$ ,

$x^3+y^3+z^3=3xyz=18\Rightarrow xyz=6$

Now (according to me) after checking the possible cases $x=3,y=-2$ and $z=-1$ .

For cross-checking, $x^7+y^7+z^7=(3)^7+(-2)^7+(-1)^7=2058$

Therefore, $x^{10}+y^{10}+z^{10}=(3)^{10}+(-2)^{10}+(-1)^{10}=\boxed{60074}$