Đáp án:

$\begin{array}{l}
a){x^6} - 7{x^3} - 8 = 0\\
 \Rightarrow {\left( {{x^3}} \right)^2} - 8{x^3} + {x^3} - 8 = 0\\
 \Rightarrow \left( {{x^3} - 8} \right).{x^3} + {x^3} - 8 = 0\\
 \Rightarrow \left( {{x^3} - 8} \right)\left( {{x^3} + 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
{x^3} = 8\\
{x^3} =  - 1
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
\text{Vậy}\,x = 2;x =  - 1\\
b){x^3} - 7x + 6 = 0\\
 \Rightarrow {x^3} - {x^2} + {x^2} - x - 6x + 6 = 0\\
 \Rightarrow {x^2}\left( {x - 1} \right) + x\left( {x - 1} \right) - 6\left( {x - 1} \right) = 0\\
 \Rightarrow \left( {x - 1} \right)\left( {{x^2} + x - 6} \right) = 0\\
 \Rightarrow \left( {x - 1} \right)\left( {{x^2} + 3x - 2x - 6} \right) = 0\\
 \Rightarrow \left( {x - 1} \right)\left( {x + 3} \right)\left( {x - 2} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 3\\
x = 2
\end{array} \right.\\
\text{Vậy}\,x = 1;x =  - 3;x = 2\\
c){\left( {{x^2} + x} \right)^2} + 4\left( {{x^2} + x} \right) = 12\\
\text{Đặt}:{x^2} + x = a\\
 \Rightarrow {a^2} + 4a = 12\\
 \Rightarrow {a^2} + 4a - 12 = 0\\
 \Rightarrow {a^2} - 2a + 6a - 12 = 0\\
 \Rightarrow \left( {a - 2} \right)\left( {a + 6} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
a - 2 = 0\\
a + 6 = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
{x^2} + x - 2 = 0\\
{x^2} + x + 6 = 0\left( {vn} \right)
\end{array} \right.\\
 \Rightarrow {x^2} + 2x - x - 2 = 0\\
 \Rightarrow \left( {x + 2} \right)\left( {x - 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x =  - 2\\
x = 1
\end{array} \right.\\
\text{Vậy}\,x =  - 2;x = 1\\
d)x\left( {x - 2} \right)\left( {x - 1} \right)\left( {x + 1} \right) = 24\\
 \Rightarrow x\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 1} \right) = 24\\
 \Rightarrow \left( {{x^2} - x} \right)\left( {{x^2} - x - 2} \right) = 24\\
\text{Đặt}:{x^2} - x = a\\
 \Rightarrow a\left( {a - 2} \right) = 24\\
 \Rightarrow {a^2} - 2a - 24 = 0\\
 \Rightarrow {a^2} - 6a + 4a - 24 = 0\\
 \Rightarrow \left( {a - 6} \right)\left( {a + 4} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
a - 6 = 0\\
a + 4 = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
{x^2} - x - 6 = 0\\
{x^2} - x + 4 = 0\left( {vn} \right)
\end{array} \right.\\
 \Rightarrow \left( {x - 3} \right)\left( {x + 2} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 3\\
x =  - 2
\end{array} \right.\\
\text{Vậy}\,x = 3;x =  - 2
\end{array}$