Resposta:   O valor da expressão

    [tex]\dfrac{4}{x^4+y^4+z^4}[/tex]

é igual a 4/33.

Explicação passo a passo:

Sejam

     [tex]\begin{array}{lc}A=x+y+z=1&\quad\mathrm{(i)}\\\\ B=x^2+y^2+z^2=9&\quad\mathrm{(ii)}\\\\ C=x^3+y^3+z^3=1&\quad\mathrm{(iii)}\end{array}[/tex]

Desenvolvendo [tex]A^2,[/tex] temos

    [tex]\begin{array}{rl}A^2=&(x+y+z)^2\\\\ =&x^2+2x(y+z)+(y+z)^2\\\\ =&x^2+2x(y+z)+(y^2+2yz+z^2)\\\\ =&x^2+2xy+2xz+y^2+2yz+z^2\\\\ =&(x^2+y^2+z^2)+2(xy+xz+yz)\\\\ =&B+2(xy+xz+yz)\end{array}\\\\\\ \begin{array}{rl}\Longleftrightarrow\quad xy+xz+yz=&\dfrac{A^2-B}{2}\\\\ =&\dfrac{1^2-9}{2}\\\\ =&-4\end{array}[/tex]

Fazendo [tex]xy+yz+xz=D,[/tex] temos

    [tex]D=xy+xz+yz=-4\qquad\mathrm{(iv)}[/tex]

Multiplicando (i) e (iv) membro a membro, obtemos

    [tex](x+y+z)(xy+xz+yz)=1\cdot (-4)\\\\ \Longleftrightarrow\quad x^2y+x^2z+xyz+xy^2+xyz+y^2z+xyz+xz^2+yz^2=-4\\\\ \Longleftrightarrow\quad (x^2y+x^2z+xy^2+y^2z+xz^2+yz^2)+3(xyz)=-4[/tex]

Fazendo

    [tex]\begin{array}{lc}E=x^2y+x^2z+xy^2+y^2z+xz^2+yz^2&\quad\mathrm{(v)}\\\\ F=xyz&\quad\mathrm{(vi)}\end{array}[/tex]

temos

    [tex]\Longrightarrow\quad E+3F=-4\qquad\mathrm{(vii)}[/tex]

Por outro lado, desenvolvendo [tex]A^3,[/tex] temos

    [tex]\begin{array}{rl}A^3=&(x+y+z)^3\\\\ =&x^3+3x^2(y+z)+3x(y+z)^2+(y+z)^3\\\\ =&x^3+3x^2(y+z)+3x(y^2+2yz+z^2)+(y^3+3y^2z+3yz^2+z^3)\\\\ =&x^3+3x^2y+3x^2z+3xy^2+6xyz+3xz^2+y^3+3y^2z+3yz^2+z^3\\\\ =&(x^3+y^3+z^3)+3(x^2y+x^2z+xy^2+xz^2+y^2z+yz^2)+6(xyz)\\\\ =&C+3E+6F\end{array}[/tex]

    [tex]\begin{array}{ll}\Longleftrightarrow\quad 3E+6F=&A^3-C\\\\ \Longleftrightarrow\quad 3(E+2F)=&A^3-C\\\\ \Longleftrightarrow\quad E+2F=&\dfrac{A^3-C}{3}\\\\ \Longrightarrow\quad E+2F=&\dfrac{1^3-1}{3}\\\\\ \Longleftrightarrow\quad E+2F=&0\qquad\mathrm{(viii)}\end{array}[/tex]

Resolva o sistema linear formado pelas equações (vii) e (viii):

    [tex]\left\{\begin{array}{l}E+3F=-4\\\\ E+2F=0 \end{array}\right.[/tex]

Isolando E na segunda equação e substituindo na primeira:

    [tex]E=-2F\\\\\\ \Longrightarrow\quad -2F+3F=-4\\\\ \Longleftrightarrow\quad F=-4\qquad\mathrm{(ix)}[/tex]

segue que

    [tex]\Longrightarrow\quad E=-2\cdot (-4)\\\\ \Longleftrightarrow\quad E=8\qquad\mathrm{(x)}[/tex]

Por (iv), desenvolvendo [tex]D^2,[/tex] temos

    [tex]\begin{array}{rl}D^2=&(xy+xz+yz)^2\\\\ =&(xy)^2+2(xy)(xz+yz)+(xz+yz)^2\\\\ =&x^2y^2+2xy(xz+yz)+[(xz)^2+2(xz)(yz)+(yz)^2]\\\\ =&x^2y^2+2x^2yz+2xy^2z+x^2z^2+2xyz^2+y^2z^2\\\\ =&(x^2y^2+x^2z^2+y^2z^2)+2xyz(x+y+z)\\\\ =&(x^2y^2+x^2z^2+y^2z^2)+2FA\end{array}\\\\\\ \begin{array}{rl}\Longleftrightarrow\quad x^2y^2+x^2z^2+y^2z^2=&D^2-2FA\\\\ =&(-4)^2-2\cdot (-4)\cdot (1)\\\\ \quad =&16+8\\\\ =&24\end{array}[/tex]

Fazendo [tex]G=x^2y^2+x^2z^2+y^2z^2,[/tex] temos

    [tex]G=x^2y^2+x^2z^2+y^2z^2=24\qquad\mathrm{(xi)}[/tex]

Por (ii), desenvolvendo [tex]B^2,[/tex] temos

    [tex]\begin{array}{rl}B^2=&(x^2+y^2+z^2)^2\\\\ =&(x^2)^2+2x^2(y^2+z^2)+(y^2+z^2)^2\\\\ =&x^4+2x^2(y^2+z^2)+[(y^2)^2+2y^2z^2+(z^2)^2]\\\\ =&x^4+2x^2y^2+2x^2z^2+y^4+2y^2z^2+z^4\\\\ =&(x^4+y^4+z^4)+2(x^2y^2+x^2z^2+y^2z^2)\\\\ =&(x^4+y^4+z^4)+2G\end{array}\\\\\\ \begin{array}{rl}\Longleftrightarrow\quad x^4+y^4+z^4=&B^2-2G\\\\ =&9^2-2\cdot 24\\\\ =&81-48\\\\ =&33\end{array}[/tex]

Portanto, o valor da expressão pedida é

    [tex]\Longrightarrow\quad \dfrac{4}{x^4+y^4+z^4}=\dfrac{4}{33}[/tex]

sendo esta a resposta.

Dúvidas? Comente.

Bons estudos! :-)