\[\begin{array}{l}
{(4t + 2)^3} + 8{(1 - 2t)^3}\\
= {\left[ {2\left( {2t + 1} \right)} \right]^3} + 8{\left( {1 - 2t} \right)^3}\\
= 8{\left( {2t + 1} \right)^3} + 8{\left( {1 - 2t} \right)^3}\\
= 8\left( {2t + 1 + 1 - 2t} \right)\left[ {{{\left( {2t + 1} \right)}^2} - \left( {2t + 1} \right)\left( {1 - 2t} \right) + {{\left( {1 - 2t} \right)}^2}} \right]\\
= 8.2.\left( {4{t^2} + 4t + 1 - 1 + 4{t^2} + 1 - 4t + 4{t^2}} \right)\\
= 16\left( {12{t^2} + 1} \right)\\
{x^3} + {y^3} - {z^3} + 3xyz\\
= {\left( {x + y} \right)^3} - 3xy\left( {x + y} \right) - {z^3} + 3xyz\\
= {\left( {x + y - z} \right)^3} + 3\left( {x + y} \right)z\left( {x + y - z} \right) - 3xy\left( {x + y - z} \right)\\
= \left( {x + y - z} \right)\left[ {{{\left( {x + y - z} \right)}^2} + 3\left( {x + y} \right)z - 3xy} \right]\\
= \left( {x + y - z} \right)\left( {{x^2} + {y^2} + {z^2} + 2xy - 2yz - 2zx + 3xz + 3yz - 3xy} \right)\\
= \left( {x + y - z} \right)\left( {{x^2} + {y^2} + {z^2} - xy + yz + zx} \right)\\
{x^2} + 6x + 8\\
= \left( {{x^2} + 6x + 9} \right) - 1\\
= {\left( {x + 3} \right)^2} - 1\\
= \left( {x + 3 - 1} \right)\left( {x + 3 + 1} \right)\\
= \left( {x + 2} \right)\left( {x + 4} \right)\\
2{x^2} + 14x + 12\\
= 2\left( {{x^2} + 7x + \frac{{49}}{4}} \right) - \frac{{25}}{2}\\
= 2{\left( {x + \frac{7}{2}} \right)^2} - 2.{\left( {\frac{5}{2}} \right)^2}\\
= 2\left( {x + \frac{7}{2} - \frac{5}{2}} \right)\left( {x + \frac{7}{2} + \frac{5}{2}} \right)\\
= 2\left( {x - 1} \right)\left( {x + 6} \right)\\
9{x^2} + 24x + 15\\
= {\left( {3x} \right)^2} + 2.3x.4 + 16 - 1\\
= {\left( {3x + 4} \right)^2} - {1^2}\\
= \left( {3x + 4 - 1} \right)\left( {3x + 4 + 1} \right)\\
= \left( {3x + 3} \right)\left( {3x + 5} \right)\\
= 3\left( {x + 1} \right)\left( {3x + 5} \right)\\
6{x^2} - xy - 7{y^2}\\
= 6\left( {{x^2} - \frac{{xy}}{6} + \frac{{{y^2}}}{{144}}} \right) - \frac{{169}}{{24}}{y^2}\\
= 6{\left( {x - \frac{y}{{12}}} \right)^2} - 6.{\left( {\frac{{13}}{{12}}y} \right)^2}\\
= 6\left( {x - \frac{y}{{12}} - \frac{{13}}{{12}}y} \right)\left( {x - \frac{y}{{12}} + \frac{{13}}{{12}}y} \right)\\
= 6\left( {x - \frac{7}{6}y} \right)\left( {x + y} \right)
\end{array}\]