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Question Number 167566 by infinityaction last updated on 19/Mar/22

Answered by MJS_new last updated on 19/Mar/22

$$x=u-v\wedge y=u+v$$$$z=1-2u$$$$z^2=1-2(u^2+v^2)$$$$\Rightarrow v^2=-u(3u-2)$$$$\Rightarrow x^3+y^3+z^3=-24u^3+24u^2-6u+1$$$$\min (-24u^3+24u^2-6u+1)=-\mathrm{\infty }$$

Commented by infinityaction last updated on 19/Mar/22

$$no$$$$answeris5/9$$

Commented by MJS_new last updated on 19/Mar/22

$$\mathrm{example}$$$$\mathrm{u}=2$$$$\Rightarrow$$$$x=2-2\sqrt{2}\mathrm{i}$$$$y=2+2\sqrt{2}\mathrm{i}$$$$z=-3$$$$x^3+y^3+z^3=-107$$$$$$$$\mathrm{we}\mathrm{can}\mathrm{solve}$$$$-24u^3+24u^2-6u+1=r$$$$\mathrm{for}\mathrm{any}r\in \mathbb{R}$$$$\mathrm{or}\mathrm{we}\mathrm{can}\mathrm{choose}\mathrm{any}\mathrm{value}u\in \mathbb{R}$$$$$$$$\mathrm{only}\mathrm{if}\mathrm{you}\mathrm{restrict}:x,y,z\in \mathbb{R}$$$$\Rightarrow -u(3u-2)⩾0\iff 0⩽u⩽\frac{2}{3}$$$$\Rightarrow \frac{5}{9}⩽r⩽1$$$$r\mathrm{has}\mathrm{local}\mathrm{extremes}\mathrm{at}u=\frac{1}{6}\vee u=\frac{1}{2}$$

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