%@P:exocorcp %@Dif:4 On donne : \[\Eqalign{ A&=(x+y+z)^2-(-x+y+z)^2\cr B&=(x+y-z)^2-(x-y+z)^2\cr C&=A-B\cr }\] \begin{myenumerate} \item Factorise les expressions $A$ et $B$ ; \item Factorise $C$. \end{myenumerate} %@Correction: \begin{myenumerate} \item \[\Eqalign{ A&=(x+y+z)^2-(-x+y+z)^2&B&=(x+y-z)^2-(x-y+z)^2\cr A&=\left[(x+y+z)-(-x+y+z)\right]\times\left[(x+y+z)+(-x+y+z)\right]&B&=\left[(x+y-z)-(x-y+z)\right]\times\left[(x+y-z)+(x-y+z)\right]\cr A&=(x+y+z+x-y-z)\times(x+y+z-x+y+z)&B&=(x+y-z-x+y-z)\times(x+y-z+x-y+z)\cr A&=2x\times(2y+2z)&B&=(2y-2z)\times2x\cr }\] \item \[\Eqalign{ C&=A-B\cr C&=2x(2y+2z)-2x(2y-2z)\cr C&=2x\times\left[(2y+2z)-(2y-2z)\right] C&=2x(2y+2z-2y+2z)\cr C&=2x\times4z\cr C&=8xz\cr }\] \end{myenumerate} %@Commentaire: Exercice assez délicat, surtout la 1\iere\ question. \`A donner en approfondissement. Travail sur la factorisation $a^2-b^2$.