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Question

# Show that: $2\left({\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x\right)-3\left({\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x\right)+1=0$

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Solution

## $\mathrm{LHS}=2\left({\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x\right)-3\left({\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x\right)+1\phantom{\rule{0ex}{0ex}}=2\left\{{\left({\mathrm{sin}}^{2}x\right)}^{3}+{\left({\mathrm{cos}}^{2}\right)}^{3}\right\}-3\left({\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x\right)+1\phantom{\rule{0ex}{0ex}}$ $=2\left[\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x\right)\left({\mathrm{sin}}^{4}x-{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{4}x\right)\right]-3\left({\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x\right)+1\phantom{\rule{0ex}{0ex}}=2\left({\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x\right)-2{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x-3\left({\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x\right)+1\phantom{\rule{0ex}{0ex}}=-\left[{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x+2{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x\right]+1\phantom{\rule{0ex}{0ex}}=-1+1=0=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}.$

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