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Question Number 96530 by abony1303 last updated on 02/Jun/20

If: (1+m)^3 −3m^3 =2        0≤m≤1  Find: (1+m)^2 −3m^2

$$\boldsymbol{\mathrm{If}}:\:\left(\mathrm{1}+{m}\right)^{\mathrm{3}} −\mathrm{3}{m}^{\mathrm{3}} =\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{0}\leqslant{m}\leqslant\mathrm{1} \\ $$$$\boldsymbol{\mathrm{Find}}:\:\left(\mathrm{1}+{m}\right)^{\mathrm{2}} −\mathrm{3}{m}^{\mathrm{2}} \\ $$

Commented by MJS last updated on 02/Jun/20

I don′t think there′s a “nice” solution  m=(1/2)−(√3)sin ((arcsin ((2(√3))/9))/3)  ⇒  (1+m)^2 −3m^2 =−(3/2)+3cos ((2arcsin ((2(√3))/9))/3)

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{there}'\mathrm{s}\:\mathrm{a}\:``\mathrm{nice}''\:\mathrm{solution} \\ $$$${m}=\frac{\mathrm{1}}{\mathrm{2}}−\sqrt{\mathrm{3}}\mathrm{sin}\:\frac{\mathrm{arcsin}\:\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}}{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}+{m}\right)^{\mathrm{2}} −\mathrm{3}{m}^{\mathrm{2}} =−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{3cos}\:\frac{\mathrm{2arcsin}\:\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}}{\mathrm{3}} \\ $$

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