Đáp án:

$\begin{array}{l}
a)\left( {5x - 3} \right)\left( {2x - c} \right) = a.{x^2} + b.x + 21\\
 \Leftrightarrow 5x.2x - 5x.c - 6x + 3c = a.{x^2} + b.x + 21\\
 \Leftrightarrow 10{x^2} + \left( { - 5c - 6} \right).x + 3c = a.{x^2} + b.x + 21\\
 \Leftrightarrow \left\{ \begin{array}{l}
a = 10\\
 - 5c - 6 = b\\
21 = 3c
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
a = 10\\
b =  - 5c - 6\\
c = 7
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
a = 10\\
b =  - 5.7 - 6 =  - 41\\
c = 7
\end{array} \right.\\
Vậy\,a = 10;b =  - 41;c = 7\\
b)\\
\left( {a.x + 4} \right)\left( {{x^2} + b.x - 1} \right) = 9{x^3} + 58{x^2} + 15x + c\\
 \Leftrightarrow a.{x^3} + a.b.{x^2} - a.x + 4{x^2} + 4bx - 4\\
 = 9{x^3} + 58{x^2} + 15x + c\\
 \Leftrightarrow a.{x^3} + \left( {a.b + 4} \right).{x^2} + \left( { - a + 4b} \right).x - 4\\
 = 9{x^3} + 58{x^2} + 15x + c\\
 \Leftrightarrow \left\{ \begin{array}{l}
a = 9\\
a.b + 4 = 58\\
 - a + 4b = 15\\
 - 4 = c
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
a = 9\\
9b = 54\\
9 + 4b = 15\\
c =  - 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
a = 9\\
b = 6\\
9 + 4.6 = 15\left( {ktm} \right)\\
c =  - 4
\end{array} \right.\\
Vậy\left( {\,a;b;c} \right) \in \emptyset \\
c)\left( {x + a} \right).\left( {{x^2} + bx + 16} \right) = {x^3} - 64\\
 \Leftrightarrow {x^3} + b.{x^2} + 16.x + a.{x^2} + a.b.x + 16a\\
 = {x^3} - 64\\
 \Leftrightarrow {x^3} + \left( {a + b} \right).{x^2} + \left( {a.b + 16} \right).x + 16.a\\
 = {x^3} - 64\\
 \Leftrightarrow \left\{ \begin{array}{l}
a + b = 0\\
a.b + 16 = 0\\
16.a =  - 64
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
b =  - a\\
a.b =  - 16\\
a =  - 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
b = 4\\
a =  - 4
\end{array} \right.\\
Vậy\,a =  - 4;b = 4
\end{array}$