Correct Answer - C
We have, `x+y+z=12" and "x^(3)y^(4)z^(5)=(0.1)(600)^(3)`.
Now,
`x+y+z=12`
`implies" "3((x)/(3))+4((y)/(4))+5((z)/(5))=12`
Now, AM of 3 identical `(x)/(3),`4 identical `(y)/(4)` and 5 identical `(z)/(5)` is
`A=(3((x)/(3))+4((y)/(4))+5((z)/(5)))/(3+4+5)=(12)/(12)=1`
G.M. of 3 identical `(x)/(3)`4 identical `(y)/(4)` and 5 identical `(z)/(5)` is
`G={((x)/(3))^(3)xx((y)/(4))^(4)xx((z)/(5))^(5)}^(1//2)=(x^(3)y^(4)z^(5))/((60)^(3)xx100)=(0.1xx(600)^(3))/((60)^(3)xx100)=1`
`:." "A=G`
`implies" "(x)/(3)=(y)/(4)=(z)/(5)`
`implies" "(x)/(3)=(y)/(4)=(z)/(5)=(x+y+z)/(12)implies(x)/(3)=(y)/(4)=(z)/(5)=1`
`implies" "x=3,y=4,z=5`
Hence, `x^(3)+y^(3)+z^(3)=27+64+125+64+125=216.`
