Đáp án:
\[P = 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
P = {\cos ^3}x.\cos 3x + {\sin ^3}x.\sin 3x + 1 - {\cos ^3}2x\\
= {\cos ^3}x.\left( {4{{\cos }^3}x - 3\cos x} \right) + {\sin ^3}x.\left( {3\sin x - 4{{\sin }^3}x} \right) + 1 - {\cos ^3}2x\\
= 4\left( {{{\cos }^6}x - {{\sin }^6}x} \right) - 3\left( {{{\cos }^4}x - {{\sin }^4}x} \right) + 1 - {\cos ^3}2x\\
= 4.\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left( {{{\cos }^4}x + {{\cos }^2}x.{{\sin }^2}x + {{\sin }^4}x} \right) - 3.\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left( {{{\cos }^2}x + {{\sin }^2}x} \right) + 1 - {\cos ^3}2x\\
= 4.\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - {{\sin }^2}x.{{\cos }^2}x} \right] - 3.\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left( {{{\cos }^2}x + {{\sin }^2}x} \right) + 1 - {\cos ^3}2x\\
= 4.\cos 2x.\left( {1 - {{\sin }^2}x.{{\cos }^2}x} \right) - 3.cos2x.1 + 1 - {\cos ^3}2x\\
= \cos 2x\left( {4 - 4{{\sin }^2}x.{{\cos }^2}x - 3 - {{\cos }^2}2x} \right) + 1\\
= \cos 2x.\left( {4 - {{\left( {2\sin x.\cos x} \right)}^2} - 3 - {{\cos }^2}2x} \right) + 1\\
= \cos 2x.\left( {1 - {{\sin }^2}2x - {{\cos }^2}2x} \right) + 1\\
= \cos 2x.0 + 1\\
= 1
\end{array}\)
Vậy \(P = 1\)
