Let x, y, z be positive real numbers such tha | vedclass

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(B) Given $x + y + z = 12$ and $x^3y^4z^5 = (0.1)(600)^3$.Using the Weighted Arithmetic Mean-Geometric Mean Inequality $(AM \ge GM)$:$\frac{3(\frac{x}

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$216$

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(B) Given $x + y + z = 12$ and $x^3y^4z^5 = (0.1)(600)^3$.Using the Weighted Arithmetic Mean-Geometric Mean Inequality $(AM \ge GM)$:$\frac{3(\frac{x}{3}) + 4(\frac{y}{4}) + 5(\frac{z}{5})}{3+4+5} \ge \sqrt[12]{(\frac{x}{3})^3 (\frac{y}{4})^4 (\frac{z}{5})^5}$$\frac{x+y+z}{12} \ge \sqrt[12]{\frac{x^3y^4z^5}{3^3 4^4 5^5}}$Since $x+y+z = 12$,we have $1 \ge \sqrt[12]{\frac{x^3y^4z^5}{3^3 4^4 5^5}}$,which implies $x^3y^4z^5 \le 3^3 4^4 5^5$.Calculating $3^3 4^4 5^5 = 27 	imes 256 	imes 3125 = 21600000 = 21.6 	imes 10^6$.Given $x^3y^4z^5 = 0.1 	imes (600)^3 = 0.1 	imes 216000000 = 21600000$.Since the equality holds,we must have $\frac{x}{3} = \frac{y}{4} = \frac{z}{5} = k$.Then $x=3k, y=4k, z=5k$. Substituting into $x+y+z=12$ gives $3k+4k+5k=12$,so $12k=12$,which means $k=1$.Thus,$x=3, y=4, z=5$.Finally,$x^3 + y^3 + z^3 = 3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216$.

Let $x, y, z$ be positive real numbers such that $x + y + z = 12$ and $x^3y^4z^5 = (0.1)(600)^3$. Then $x^3 + y^3 + z^3$ is equal to

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