Giải thích các bước giải:

$\begin{aligned}
& \text{Ta có: } x + y + z = \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \\
& \Leftrightarrow x + y + z = \dfrac{xy + yz + zx}{xyz} \\
& \text{Thay } xyz = 1: \\
& \Leftrightarrow x + y + z = xy + yz + zx \\
& \Leftrightarrow xy + yz + zx - x - y - z = 0 \\
& \text{Cộng hai vế với } xyz - 1 \text{ (vì } xyz - 1 = 0 \text{):} \\
& \Leftrightarrow (xyz - 1) - (xy + yz + zx - x - y - z) = 0 \\
& \Leftrightarrow xyz - xy - yz - zx + x + y + z - 1 = 0 \\
& \Leftrightarrow (xyz - xy) - (xz - x) - (yz - y) + (z - 1) = 0 \\
& \Leftrightarrow xy(z - 1) - x(z - 1) - y(z - 1) + (z - 1) = 0 \\
& \Leftrightarrow (z - 1)(xy - x - y + 1) = 0 \\
& \Leftrightarrow (z - 1)[x(y - 1) - (y - 1)] = 0 \\
& \Leftrightarrow (x - 1)(y - 1)(z - 1) = 0 \\
& \Leftrightarrow \left[ \begin{aligned} & x - 1 = 0 \\ & y - 1 = 0 \\ & z - 1 = 0 \end{aligned} \right. \\
& \Leftrightarrow \left[ \begin{aligned} & x = 1 \\ & y = 1 \\ & z = 1 \end{aligned} \right. \\
& \Rightarrow A = (x^{68} - 1)(y^{69} - 1)(z^{70} - 1) = 0
\end{aligned}$