Đáp án:

\(\begin{array}{l}
a)\,\,\,\left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right. &  &  & b)\,\,\left[ \begin{array}{l}
x =  - 5\\
x = 2\\
x =  - 2
\end{array} \right.\\
c)\,\,\left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right. &  &  & d)\,\,\left[ \begin{array}{l}
x =  - 5\\
x = 3
\end{array} \right.\\
e)\,\,x = 2 &  &  & g)\,\,\left[ \begin{array}{l}
x = 8\\
x =  - 2
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\,\,\,{x^3} - 8 = {\left( {x - 2} \right)^3}\\
 \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) - \left( {x - 2} \right)\left( {{x^2} - 4x + 4} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 4 - {x^2} + 4x - 4} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)6x = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
6x = 0\\
x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right..\\
b)\,\,{x^3} + 5{x^2} - 4x - 20 = 0\\
 \Leftrightarrow {x^2}\left( {x + 5} \right) - 4\left( {x + 5} \right) = 0\\
 \Leftrightarrow \left( {x + 5} \right)\left( {{x^2} - 4} \right) = 0\\
 \Leftrightarrow \left( {x + 5} \right)\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 5 = 0\\
x - 2 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 5\\
x = 2\\
x =  - 2
\end{array} \right..\\
c)\,\,{x^3} - 4{x^2} + 4x = 0\\
 \Leftrightarrow x\left( {{x^2} - 4x + 4} \right) = 0\\
 \Leftrightarrow x{\left( {x - 2} \right)^2} = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right..\\
d)\,\,{x^2} - 25 + 2\left( {x + 5} \right) = 0\\
 \Leftrightarrow \left( {x - 5} \right)\left( {x + 5} \right) + 2\left( {x + 5} \right) = 0\\
 \Leftrightarrow \left( {x + 5} \right)\left( {x - 5 + 2} \right) = 0\\
 \Leftrightarrow \left( {x + 5} \right)\left( {x - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 5 = 0\\
x - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 5\\
x = 3
\end{array} \right..\\
e)\,\,\,{x^2}\left( {x - 2} \right) + 7x = 14\\
 \Leftrightarrow {x^2}\left( {x - 2} \right) + 7x - 14 = 0\\
 \Leftrightarrow {x^2}\left( {x - 2} \right) + 7\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 7} \right) = 0\\
 \Leftrightarrow x - 2 = 0\,\,\,\,\,\left( {{x^2} + 7 > 0\,\,\,\forall x} \right)\\
 \Leftrightarrow x = 2.\\
g)\,\,{x^2} - 25 = 6x - 9\\
 \Leftrightarrow {x^2} - 6x + 9 - 25 = 0\\
 \Leftrightarrow {\left( {x - 3} \right)^2} - {5^2} = 0\\
 \Leftrightarrow \left( {x - 3 - 5} \right)\left( {x - 3 + 5} \right) = 0\\
 \Leftrightarrow \left( {x - 8} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 8 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x =  - 2
\end{array} \right..
\end{array}\)