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trigonometry equation - ${\mathrm{sin}}^{3}\left(x\right)+{\mathrm{sin}}^{3}\left(2x\right)+{\mathrm{sin}}^{3}\left(3x\right)=\left(\mathrm{sin}\left(x\right)+\mathrm{sin}\left(2x\right)+\mathrm{sin}\left(3x\right){\right)}^{3}$
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Lilliana Burton
Let $\mathrm{sin}x=a$, $\mathrm{sin}2x=b$, and $\mathrm{sin}3x=c$
Expand out
$\left(a+b+c{\right)}^{3}$
Then subtract ${a}^{3},{b}^{3}$ and ${c}^{3}$, which yields
$3{a}^{2}b+3{a}^{2}c+3a{b}^{2}+6abc+3a{c}^{2}+3{b}^{2}c+3b{c}^{2}=0$
The long expression can be factored nicely into :
3(a+b)(b+c)(a+c), which equal to 0.
The factor 3 makes no difference, so solve for:
$\mathrm{sin}\left(x\right)+\mathrm{sin}\left(2x\right)=0$,
$\mathrm{sin}\left(2x\right)+\mathrm{sin}\left(3x\right)=0$, and
$\mathrm{sin}\left(x\right)+\mathrm{sin}\left(3x\right)=0$.