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Question Number 161242 by HongKing last updated on 14/Dec/21

$$\mathrm{x};\mathrm{y};\mathrm{z}<0$$ $$\frac{{\mathrm{x}}^3}{\sqrt{\mathrm{yz}}}=-3;\frac{{\mathrm{y}}^3}{\sqrt{\mathrm{xz}}}=-6;\frac{{\mathrm{z}}^3}{\sqrt{\mathrm{xy}}}=-8$$ $$\mathrm{find}\mathrm{x}\cdot \mathrm{y}\cdot \mathrm{z}=?$$

Answered by mr W last updated on 15/Dec/21

$$\frac{x^3{y}^3{z}^3}{\sqrt{x^2{y}^2{z}^2}}=-3\times 6\times 8$$ $$\frac{x^3{y}^3{z}^3}{-xyz}=-3\times 6\times 8$$ $${\left(xyz\right)}^2=3\times 6\times 8={12}^2$$ $$\Rightarrow xyz=-12$$

Commented byHongKing last updated on 17/Dec/21

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

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