Giải thích các bước giải:

 6.6:

ĐKXĐ: $abc\ne 0$

Ta có:

$\dfrac{{{a^2}}}{{\left( {a - b} \right)\left( {a + b} \right) - {c^2}}} = \dfrac{{{a^2}}}{{{a^2} - {b^2} - {c^2}}} = \dfrac{{{a^2}}}{{{{\left( { - b - c} \right)}^2} - {b^2} - {c^2}}} = \dfrac{{{a^2}}}{{2bc}} = \dfrac{{{a^3}}}{{2abc}}$

Tương tự có:

$\dfrac{{{b^2}}}{{\left( {b - c} \right)\left( {b + c} \right) - {a^2}}} = \dfrac{{{b^3}}}{{2abc}};\dfrac{{{c^2}}}{{\left( {c - a} \right)\left( {c + a} \right) - {b^2}}} = \dfrac{{{c^3}}}{{2abc}}$

Khi đó:

$\begin{array}{l}
 \Rightarrow T = \dfrac{{{a^2}}}{{\left( {a - b} \right)\left( {a + b} \right) - {c^2}}} + \dfrac{{{b^2}}}{{\left( {b - c} \right)\left( {b + c} \right) - {a^2}}} + \dfrac{{{c^2}}}{{\left( {c - a} \right)\left( {c + a} \right) - {b^2}}}\\
 = \dfrac{{{a^3}}}{{2abc}} + \dfrac{{{b^3}}}{{2abc}} + \dfrac{{{c^3}}}{{2abc}}\\
 = \dfrac{{{a^3} + {b^3} + {c^3}}}{{2abc}}\\
 = \dfrac{{\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ac} \right) + 3abc}}{{2abc}}\\
 = \dfrac{{3abc}}{{2abc}}\left( {a + b + c = 0} \right)\\
 = \dfrac{3}{2}
\end{array}$

Vậy $T= \dfrac{3}{2}$

6.7:

ĐKXĐ: $abc\ne 0; a+b+c\ne 0$

Ta có:

$\begin{array}{l}
P = \left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)\left( {\dfrac{1}{b} + \dfrac{1}{c}} \right)\left( {\dfrac{1}{c} + \dfrac{1}{a}} \right)\\
 = \dfrac{{a + b}}{{ab}}.\dfrac{{b + c}}{{bc}}.\dfrac{{a + c}}{{ac}}\\
 = \dfrac{{\left( {a + b} \right)\left( {b + c} \right)\left( {a + c} \right)}}{{{{\left( {abc} \right)}^2}}}
\end{array}$

Lại có:

$\begin{array}{l}
{a^3} + {b^3} + {c^3} = 3abc\\
 \Leftrightarrow \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ac} \right) + 3abc = 3abc\\
 \Leftrightarrow \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ac} \right) = 0\\
 \Leftrightarrow {a^2} + {b^2} + {c^2} - ab - bc - ac = 0\left( {a + b + c \ne 0} \right)\\
 \Leftrightarrow 2\left( {{a^2} + {b^2} + {c^2} - ab - bc - ac} \right) = 0\\
 \Leftrightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {a - c} \right)^2} = 0\\
 \Leftrightarrow a - b = b - c = c - a = 0\\
 \Leftrightarrow a = b = c
\end{array}$

Như vậy:

$\begin{array}{l}
P = \dfrac{{\left( {a + b} \right)\left( {b + c} \right)\left( {a + c} \right)}}{{{{\left( {abc} \right)}^2}}}\\
 = \dfrac{{2a.2a.2a}}{{{{\left( {a.a.a} \right)}^2}}}\\
 = \dfrac{8}{{{a^3}}}\\
 = \dfrac{8}{{abc}}
\end{array}$

Ta có điều phải chứng minh.