(x+y)(x^2-y^2)+(y+z)(y^2-z^2)+(z+x)(z^2-x^2)

0 bình luận về “(x+y)(x^2-y^2)+(y+z)(y^2-z^2)+(z+x)(z^2-x^2)”

1. Đáp án
   $\begin{array}{l}
   (x + y).({x^2} – {y^2}) + (y + z).({y^2} – {z^2}) + (z + x)({z^2} – {x^2})\\
   = {x^3} – x{y^2} + {x^2}y – {y^3} + {y^3} – y{z^2} + {y^2}z – {z^3} + {z^3} – {x^2}z + x{z^2} – {x^3}\\
   = – x{y^2} + {x^2}y – y{z^2} + {y^2}z – {x^2}z + x{z^2}\\
   = (x{z^2} – x{y^2}) + ( – y{z^2} + {y^2}z) + ({x^2}y – {x^2}z)\\
   = x({z^2} – {y^2}) + yz(y – z) + {x^2}(y – z)\\
   = (y – z){\rm{[ – x(y + z) + yz + }}{{\rm{x}}^2}{\rm{]}}\\
   = (y – z){\rm{[}}{{\rm{x}}^2} + yz – xy – xz]
   \end{array}$

   Bình luận