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Question Number 190131 by HeferH last updated on 28/Mar/23

$$\mathrm{if}:{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2+14=2(\mathrm{x}+2\mathrm{y}+3\mathrm{z})$$$$\mathrm{find}:\mathrm{T}=\frac{\mathrm{xyz}}{{\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3}$$

Answered by som(math1967) last updated on 28/Mar/23

$$x^2+y^2+z^2+14-2x-4y-6z=0$$$$(x^2-2x+1)+(y^2-4y+4)+(z^2-6z+9)=0$$$${(x-1)}^2+{(y-2)}^2+{(z-3)}^2=0$$$$ifx,y,zarerealthen$$$${(x-1)}^2=0\Rightarrow x=1$$$${(y-2)}^2=0\Rightarrow y=2$$$${(z-3)}^2=0\Rightarrow z=3$$$$T=\frac{xyz}{x^3+y^3+z^3}=\frac{1.2.3}{1+8+27}=\frac{6}{36}=\frac{1}{6}$$

Commented by HeferH last updated on 28/Mar/23

$$\mathrm{thank}\mathrm{you}:)$$

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