Question

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

• A. $258$
• B. $216$
• C. $342$
• D. $270$

Answer 1

Answer: (B)

$216$

Given,

$x+y+z=12$

And, $x^3{y}^4{z}^5=60\times 600\times 600$

Factorize it

$x^3{y}^4{z}^5=6^3\times {10}^5$

$=2^3\times 3^3\times 2^5\times 5^5$

$=2^8\times 3^3\times 5^5$

$=3^3\times 4^4\times 5^5$
So, we can say that $x=3,y=4$ and $z=5$

Hence, $x^3+y^3+z^3=27+64+125=216$