Giải thích các bước giải:

 Ta có:

$B = 1 + 2 + 3 + ... + 100 = \frac{{\left( {1 + 100} \right).100}}{2} = 5050$

Để CM $A \vdots B \Leftrightarrow \left\{ \begin{array}{l}
A \vdots 50\\
A \vdots 101
\end{array} \right.\left( {5050 = 50.101;\left( {50,101} \right) = 1} \right)$

$\begin{array}{l}
 + )CM:A \vdots 50\\
A = {1^3} + {2^3} + {3^3} + ... + {100^3}\\
 \Leftrightarrow A = \left( {{1^3} + {{99}^3}} \right) + \left( {{2^3} + {{98}^3}} \right) + ... + \left( {{{49}^3} + {{51}^3}} \right) + {50^3} + {100^3}\\
 \Leftrightarrow A = \left( {1 + 99} \right)\left( {{1^2} - 1.99 + {{99}^2}} \right) + \left( {2 + 98} \right)\left( {{2^2} - 2.98 + {{98}^2}} \right) + ... + \left( {49 + 51} \right)\left( {{{49}^2} - 49.51 + {{51}^2}} \right) + {50^3} + {100^3}\\
 \Leftrightarrow A = 100\left( {{1^2} - 1.99 + {{99}^2}} \right) + 100\left( {{2^2} - 2.98 + {{98}^2}} \right) + ... + 100\left( {{{49}^2} - 49.51 + {{51}^2}} \right) + {50^3} + {100^3}\\
 \Leftrightarrow A = 100\left( {{1^2} - 1.99 + {{99}^2} + {2^2} - 2.98 + {{98}^2} + {{49}^2} - 49.51 + {{51}^2}} \right) + {50^3} + {100^3}\\
 \Rightarrow A \vdots 50\\
 + )CM:A \vdots 101\\
A = {1^3} + {2^3} + {3^3} + ... + {100^3}\\
 \Leftrightarrow A = \left( {{1^3} + {{100}^3}} \right) + \left( {{2^3} + {{99}^3}} \right) + ... + \left( {{{50}^3} + {{51}^3}} \right)\\
 \Leftrightarrow A = \left( {1 + 100} \right)\left( {{1^2} - 1.100 + {{100}^2}} \right) + \left( {2 + 99} \right)\left( {{2^2} - 2.99 + {{99}^2}} \right) + ... + \left( {50 + 51} \right)\left( {{{50}^2} - 50.51 + {{51}^2}} \right)\\
 \Leftrightarrow A = 101\left( {{1^2} - 1.100 + {{100}^2}} \right) + 101\left( {{2^2} - 2.99 + {{99}^2}} \right) + ... + 101\left( {{{50}^2} - 50.51 + {{51}^2}} \right)\\
 \Leftrightarrow A = 101\left( {{1^2} - 1.100 + {{100}^2} + {2^2} - 2.99 + {{99}^2} + {{50}^2} - 50.51 + {{51}^2}} \right)\\
 \Rightarrow A \vdots 101
\end{array}$

Như vậy ta cm được $A \vdots 5050$ hay $A \vdots B$