$\displaystyle \begin{array}{{>{\displaystyle}l}} a+b+c=0\ \\ \rightarrow \frac{1}{a} +\frac{1}{b} +\frac{1}{c} =0\ \\ \rightarrow \left(\frac{1}{a} +\frac{1}{b} +\frac{1}{c}\right)^{2} =0\ ( 1)\\ \rightarrow \frac{1}{a^{2}} +\frac{1}{b^{2}} +\frac{1}{c^{2}} +2\left(\frac{1}{ab} +\frac{1}{bc} +\frac{1}{ac}\right) =0\ \\ \rightarrow \frac{1}{a^{2}} +\frac{1}{b^{2}} +\frac{1}{c^{2}} =-2\left(\frac{1}{ab} +\frac{1}{bc} +\frac{1}{ca}\right) =0\ \\ \rightarrow \frac{1}{a^{2}} +\frac{1}{b^{2}} +\frac{1}{c^{2}} =-2\left(\frac{c+a+b}{abc}\right) =0\ \\ Vì\ a+c+b=0\ nên\ :\frac{1}{a^{2}} +\frac{1}{b^{2}} +\frac{1}{c^{2}} =0\ ( 2)\\ từ\ ( 1) \ ( 2) \ \rightarrow dpcm\ \\ Bài\ 2:\ \\ \left( x^{3} -3\right) +( x-5)^{3} +8( 4-x)^{3} =0\ \\ ( x-3+x-5)\left[( x-3)^{2} -( x-3)( x-5) +( x-5)^{2}\right] +\left[ 2( 4-x)^{3}\right] =0\\ ( x-8)\left[ x^{2} -6x+9-x^{2} +8x-15+x^{2} -10x+25\right) +( 8-x)^{3} =0\ \\ ( x-8)\left( x^{2} -8x+19\right) -( x-8)^{3} =0\\ ( x-8)\left( x^{2} -8x+19-( x-8)^{2}\right) =0\ \\ ( x-8)\left( x^{2} -8x+19-x^{2} +16x-64\right) =0\ \\ ( x-8)( 8x-45) =0\ \\ \rightarrow \left[ \begin{array}{l l} x=8 & \\ x=\frac{45}{8} & \end{array} \right.\\ \\ \end{array}$
