The exercise is lớn evaluate it. In my text book the answer is $0$I tried to factor the expression, but it got me nowhere.




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$$sin^4 x+cos^4 x=(sin^2 x+cos^2 x)^2-2sin^2xcos^2x=1-2sin^2xcos^2x$$$$sin^6 x+cos^6 x=(sin^2 x+cos^2 x)^3-3sin^2 xcos^2 x(sin^2 x+cos^2 x)=1-3sin^2xcos^2x$$

Thus your expression simplifies to

$$2-6sin^2xcos^2x-3+6sin^2xcos^2x+1$$

Which is $0$, for all $x$.


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$$2(sin^6x +cos^6x) - 3(sin^4x+cos^4x)+1=$$$$=2(sin^2x +cos^2x)(sin^4x-sin^2x cos^2 x+cos^4x) - 3(sin^4x+cos^4x)+1=$$$$=2(sin^4x-sin^2x cos^2 x+cos^4x) - 3(sin^4x+cos^4x)+1=$$$$=-2sin^2x cos^2 x -sin^4x-cos^4x+1=$$$$-(sin^2x +cos^2x)^2+1=-1+1=0$$


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$$eginalign1&=left(sin^2(x)+cos^2(x) ight)^3\&=sin^6(x)+3sin^4(x)cos^2(x)+3sin^2(x)cos^4(x)+cos^6(x)\&=sin^6(x)+3sin^2(x)cos^2(x)+cos^6(x) ag1\\1&=left(sin^2(x)+cos^2(x) ight)^2\&=sin^4(x)+2sin^2(x)cos^2(x)+cos^4(x) ag2endalign$$Subtracting $3$ times $(2)$ from $2$ times $(1)$ gives $-1$. Therefore,$$2left(sin^6(x)+cos^6(x) ight)-3left(sin^4(x)+cos^4(x) ight)+1=0 ag3$$


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Consider $a^3+b^3=(a+b)^3-3ab(a+b)$ and $a^2+b^2=(a+b)^2-2ab$. For $a=sin^2x$ & $b=cos^2x$ we have$$2(sin^6x+cos^6x)-3(sin^4x+cos^4x)+1=2(a+b)^3-6ab(a+b)-3(a+b)^2+6ab+1$$However, $a+b=sin^2x+cos^2x=1$, so the expression simplifies to$$2-6ab-3+6ab+1=0$$


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