x-y-z-are-real-numbers-x-y-z-4-x-2-y-2-z-2-10-x-3-y-3-z-3-22-Find-x-4-y-4-z-4-

Question Number 120952 by abony1303 last updated on 04/Nov/20

x, y, z a r e r e a l n u m b e r s

{\begin{cases} x + y + z = 4 \\ x^{2} + y^{2} + z^{2} = 10 \\ x^{3} + y^{3} + z^{3} = 22 \end{cases} F i n d x^{4} + y^{4} + z^{4}

Answered by MJS_new last updated on 04/Nov/20

z = 4 - x - y

let x = u - \sqrt{v} \land y = u + \sqrt{v}

\Rightarrow

{\begin{cases} z = 4 - 2 u \\ 2 v + 2 (3 u^{2} - 8 u + 3) = 0 \\ 6 u v - 6 (u^{3} - 8 u^{2} + 16 u - 7) = 0 \end{cases}

\Rightarrow - 3 u^{2} - 8 u + 3 = \frac{u^{3} - 8 u^{2} + 16 u - 7}{u}

\Leftrightarrow u^{3} - 4 u^{2} + \frac{19}{4} u - \frac{7}{4} = 0

\Leftrightarrow (u - 1) (u - \frac{3 - \sqrt{2}}{2}) (u - \frac{3 + \sqrt{2}}{2}) = 0

take u = 1 [or any other of the solutions]

\Rightarrow v = 2

\Rightarrow x = 1 - \sqrt{2} \land y = 1 + \sqrt{2} \land z = 2

\Rightarrow answer is 50

Answered by Jamshidbek2311 last updated on 04/Nov/20

x y + y z + x z = 3

x y z = - 2

{(x y)}^{2} + {(y z)}^{2} + {(x z)}^{2} = 25

x^{4} + y^{4} + z^{4} = 50