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Question Number 165545 by HongKing last updated on 03/Feb/22

$$\mathrm{If}$$$${13\mathrm{x}}^2+{5\mathrm{y}}^2+{9\mathrm{z}}^2+1=4\mathrm{x}-6\mathrm{xy}-12\mathrm{yz}$$$$\mathrm{Find}$$$$(\mathrm{x}+\mathrm{y}+\mathrm{z})(\mathrm{xy}+\mathrm{xz}+\mathrm{yz})=?$$

Commented by MJS_new last updated on 04/Feb/22

$$\mathrm{if}x,y,z\in \mathbb{C}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{unique}:\mathrm{we}\mathrm{can}\mathrm{choose}$$$$\mathrm{two}\mathrm{variables}\mathrm{and}\mathrm{solve}\mathrm{for}\mathrm{the}\mathrm{third}\mathrm{one}.$$

Commented by som(math1967) last updated on 04/Feb/22

$$yessir$$

Commented by HongKing last updated on 06/Feb/22

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{Sir}$$

Answered by som(math1967) last updated on 03/Feb/22

$$(4x^2-4x+1)+(9x^2+6xy+y^2)$$$$+(4y^2+12yz+9z^2)=0$$$${(2x-1)}^2+{(3x+y)}^2+{(2y+3z)}^2=0$$$$\Rightarrow {(2x-1)}^2=0\Rightarrow x=\frac{1}{2}$$$${(3x+y)}^2=0\Rightarrow y=-3x\Rightarrow y=-\frac{3}{2}$$$${(2y+3z)}^2=0$$$$\Rightarrow 2y+3z=0\Rightarrow z=-\frac{2y}{3}=1$$$$(x+y+z)(xy+xz+yz)$$$$=(\frac{1}{2}-\frac{3}{2}+1)(xy+yz+zx)$$$$=0$$

Commented by HongKing last updated on 06/Feb/22

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{Sir}$$

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