Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
3\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - 2\left( {{{\sin }^6}x + {{\cos }^6}x} \right)\\
 = 3\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x.{{\cos }^2}x} \right] - 2.\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^3} - 3{{\sin }^2}x.{{\cos }^2}x.\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right]\\
 = 3.\left[ {{1^2} - 2{{\sin }^2}x.{{\cos }^2}x} \right] - 2.\left[ {{1^3} - 3{{\sin }^2}x.{{\cos }^2}x.1} \right]\\
 = 3 - 6{\sin ^2}x.{\cos ^2}x - 2 + 6{\sin ^2}x.{\cos ^2}x\\
 = 1\\
c,\\
\left( {{{\sin }^4}x + {{\cos }^4}x - 1} \right)\left( {{{\tan }^2}x + {{\cot }^2}x + 2} \right)\\
 = \left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x.{{\cos }^2}x - 1} \right].\left( {\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}} + 2} \right)\\
 = \left[ {{1^2} - 2{{\sin }^2}x.{{\cos }^2}x - 1} \right].\left[ {\dfrac{{{{\sin }^4}x + {{\cos }^4}x + 2{{\sin }^2}x.{{\cos }^2}x}}{{{{\sin }^2}x.{{\cos }^2}x}}} \right]\\
 =  - 2{\sin ^2}x.{\cos ^2}x.\dfrac{{{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2}}}{{{{\sin }^2}x.{{\cos }^2}x}}\\
 =  - 2{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} =  - 2.1 =  - 2\\
d,\\
{\cos ^2}x.{\cot ^2}x + 3{\cos ^2}x - {\cot ^2}x + 2{\sin ^2}x\\
 = {\cot ^2}x\left( {{{\cos }^2}x - 1} \right) + 2\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + {\cos ^2}x\\
 = \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}.\left( { - {{\sin }^2}x} \right) + 2.1 + {\cos ^2}x\\
 =  - {\cos ^2}x + 2 + {\cos ^2}x\\
 = 2\\
e,\\
\dfrac{{{{\sin }^4}x + 3{{\cos }^4}x - 1}}{{{{\sin }^6}x + {{\cos }^6}x + 3{{\cos }^4}x - 1}}\\
 = \dfrac{{\left( {{{\sin }^4}x + {{\cos }^4}x} \right) + 2{{\cos }^4}x - 1}}{{\left( {{{\sin }^6}x + {{\cos }^6}x} \right) + 3{{\cos }^4}x - 1}}\\
 = \dfrac{{{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x.{{\cos }^2}x + 2{{\cos }^4}x - 1}}{{{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^3} - 3{{\sin }^2}x.{{\cos }^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + 3{{\cos }^4}x - 1}}\\
 = \dfrac{{{1^2} + 2{{\cos }^2}x\left( {{{\cos }^2}x - {{\sin }^2}x} \right) - 1}}{{{1^3} - 3{{\sin }^2}x.{{\cos }^2}x.1 + 3{{\cos }^4}x - 1}}\\
 = \dfrac{{2{{\cos }^2}x\left( {{{\cos }^2}x - {{\sin }^2}x} \right)}}{{3{{\cos }^2}x\left( {{{\cos }^2}x - {{\sin }^2}x} \right)}}\\
 = \dfrac{2}{3}
\end{array}\)