Chứng minh $: sin^{6}x+cos^{6}x=1-3sin^{2}xcos^{2}x$
Đi thi gặp bài có công thức thế chả ai biến đổi như bạn trên cả

Áp dụng $a^3 +b^3 = (a+b)^3 - 3ab(a+b)$ có luôn và ngay


$(\sin^2 x)^3 +(\cos^2 x)^3= (\sin^2 x +\cos^2 x)^3 -3\sin^2 x \cos^2 x (\sin^2 x+\cos^2 x) = 1 -3\sin^2 x \cos^2 x$
Anh ghi nham (1 b)^3. Nen sua lai y ma :d –  Nero 23-04-14 09:03 AM
e thằng ku kia bay sưa cái j bài ta thế hả –  Dép Lê Con Nhà Quê 22-04-14 10:35 AM
Ta có  $sin^{2}x+cos^{2}x=1\Leftrightarrow (sin^{2}x+cos^{2}x) ^{3}=1                                                              \Leftrightarrow sin^{6}x+cos^{6}x+3cos^{2}x.sin^{2}x(sin^{2}x+cos^{2}x)=1                                         \Leftrightarrow sin^{6}x+cos^{6}x=1-3sin^{2}x.cos^{2}x$

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