1)
\(\begin{array}{l}
a){\left( {2x - 8} \right)^2} - 6 = 0\\
\Leftrightarrow {\left( {2x - 8} \right)^2} = 6\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 8 = \sqrt 6 \\
2x - 8 = - \sqrt 6
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{8 + \sqrt 6 }}{2}\\
x = \frac{{8 - \sqrt 6 }}{2}
\end{array} \right.\\
b)\, \Leftrightarrow 4{x^2} - 4x + 1 - \left( {4{x^2} - 11x - 3} \right) = 3\\
\Leftrightarrow 7x = 5 \Leftrightarrow x = \frac{5}{7}
\end{array}\)
2)
Ta có \({\left( {x + y} \right)^3} = {x^3} + 3{x^2}y + 3x{y^2} + {y^3} = {x^3} + {y^3} + 3xy\left( {x + y} \right)\)\( \Rightarrow {x^3} + {y^3} = {\left( {x + y} \right)^3} - 3xy\left( {x + y} \right)\)
Xét \(y = {x^3} + {y^3} + {z^3} - 3xyz\)\( = {\left( {x + y} \right)^3} - 3xy\left( {x + y} \right) + {z^3} - 3xyz\)\( = \left[ {{{\left( {x + y} \right)}^3} + {z^3}} \right] - 3xy\left( {x + y + z} \right)\)
\( = \left( {x + y + z} \right)\left[ {{{\left( {x + y} \right)}^2} - \left( {x + y} \right)z + {z^2}} \right] - 3xy\left( {x + y + z} \right)\)
Mà \(x + y + z = 0\) nên \(y = 0.\left[ {{{\left( {x + y} \right)}^2} - \left( {x + y} \right)z + {z^2}} \right] - 3xy.0 = 0\)
Vậy \(y = 0\) .
Vậy \(x^3+y^3+z^3=3xyz\)
