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Question Number 48540 by behi83417@gmail.com last updated on 25/Nov/18

Answered by ajfour last updated on 25/Nov/18

$$x^2+y^2+z^2+2(xy+yz+zx)=0$$$$\Rightarrow xy+yz+zx=-\frac{ab}{2}$$$${(x+y+z)}^3=x^3+y^3+z^3+3(y+z)x^2$$$$+3(z+x)y^2+3(x+y)z^3$$$$+6xyz$$$$nowsincex+y+z=0,$$$$\Rightarrow 0=-2\mathrm{\Sigma }{x}^3+6xyz$$$$\Rightarrow xyz=\frac{a^3+b^3}{3}$$$$\Rightarrow x,y,zarerootsofeq.$$$$t^3-\frac{ab}{2}t-\frac{(a^3+b^3)}{3}=0$$$$Trigonometricsolutiontothe$$$$abovecubicgivesx,y,andz.$$

Commented by behi83417@gmail.com last updated on 25/Nov/18

$$thankyousomuchsirAjfour.but$$$$stillwaitingforanotherquestion....$$

Commented by MJS last updated on 25/Nov/18

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{we}\mathrm{get}3\mathrm{real}\mathrm{solutions}$$$$\mathrm{because}D=\frac{p^3}{27}+\frac{q^2}{4}=\frac{6a^6+11a^3{b}^3+6b^6}{216}>0\mathrm{\forall }a,b\in \mathbb{R}$$$$\Rightarrow {\mathrm{Cardano}}^{\prime }\mathrm{s}\mathrm{solution}\Rightarrow \mathrm{only}\mathrm{one}\mathrm{variable}$$$$\mathrm{of}x,y,z\mathrm{is}\mathrm{real},\mathrm{the}\mathrm{others}\mathrm{are}\mathrm{complex}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18

$${(x+y+z)}^2=ab+2(xy+yz+zx)$$$$ab+2(xy+yz+zx)=0$$$$x^3+y^3+z^3-3xyz+3xyz=a^3+b^3$$$$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz=a^3+b^3$$$$xyz=\frac{a^3+b^3}{3}$$$$xy+yz+zx=\frac{-ab}{2}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18

$$waitpls$$

Commented by behi83417@gmail.com last updated on 25/Nov/18

$$thankyouverymuchsirtanmay.waiting.....$$

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