Đáp án:
$\begin{array}{l}
a)16{x^3}y + \frac{1}{4}y{z^3}\\
= 2y\left( {8{x^3} + \frac{1}{8}{z^3}} \right)\\
= 2y\left( {2x + \frac{1}{2}z} \right)\left( {4{x^2} - xz + \frac{1}{4}{z^2}} \right)\\
b){x^4} - {x^2} + 2x - 1\\
= {x^4} - \left( {{x^2} - 2x + 1} \right)\\
= {x^4} - {\left( {x - 1} \right)^2}\\
= \left( {{x^2} + x - 1} \right)\left( {{x^2} - x + 1} \right)\\
c){x^4} + 2{x^3} - 6x - 9 = 0\\
\Leftrightarrow {x^4} - 9 + 2{x^3} - 6x = 0\\
\Leftrightarrow \left( {{x^2} - 3} \right)\left( {{x^2} + 3} \right) - 2x\left( {{x^2} - 3} \right) = 0\\
\Leftrightarrow \left( {{x^2} - 3} \right)\left( {{x^2} + 3 - 2x} \right) = 0\\
\Leftrightarrow \left( {x - \sqrt 3 } \right)\left( {x + \sqrt 3 } \right)\left( {x - 1} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \sqrt 3 \\
x = - \sqrt 3 \\
x = 1\\
x = - 3
\end{array} \right.\\
Vậy\,x \in \left\{ { - 3; - \sqrt 3 ;1;\sqrt 3 } \right\}
\end{array}$
