$3(\sin\text{x}-\cos\text{x})^4+6(\sin\text{x}+\cos\text{x})^2+4(\sin^6\text{x}+\cos^6\text{x})=$ 13. Solution: Given expression is $3(\sin\text{x}-\cos\text{x})^4+6(\sin\text{x}+\cos\text{x})^2+4(\sin^6\text{x}+\cos^6\text{x})$ $$$=3[\sin^2\text{x}+\cos^2\text{x}-2\sin\text{x}\cos\text{x}]^2+6(\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x})\\+4[(\sin^2\text{x})^3+(\cos^2\text{x})^3]$ $=3[1-2\sin\text{x}\cos\text{x}]^2+6(1+2\sin\text{x}\cos\text{x})+4[(\sin^2\text{x}+\cos^2\text{x})^3\\-3\sin^2\text{x}\cos^2\text{x}(\sin^2\text{x}+\cos^2\text{x})]$ $=3(1+4\sin^2\text{x}\cos^2\text{x}-4\sin\text{x}\cos\text{x})+6+12\sin\text{x}\cos\text{x}+4-12\sin^2\text{x}\cos^2\text{x}$ $=3+12\sin^2\text{x}\cos^2\text{x}-12\sin\text{x}\cos\text{x}+6+12\sin\text{x}\cos\text{x}+4-12\sin^2\text{x}\cos^2\text{x}$ $=3+6+4=13$ Hence, the value of the filler is $13$.$$
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