Mathematics - Sequence & Series Question with Solution | TestHub

$x+y+z=12$

$\mathrm{A}.\mathrm{M}.\ge \mathrm{G}.\mathrm{M}.$

$\frac{3\left(\frac{x}{3}\right)+4\left(\frac{y}{4}\right)+5\left(\frac{z}{5}\right)}{12} \ge \sqrt[12]{{\left(\frac{x}{3}\right)}^3 {\left(\frac{y}{4}\right)}^4 {\left(\frac{z}{5}\right)}^5}$

$1\ge 1$

$\Rightarrow \mathrm{A}.\mathrm{M}.=\mathrm{G}.\mathrm{M}.$

$\Rightarrow$All the numbers are equal.

$\therefore$$\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\left(=k\right)$

$x=3k; y=4k; z=5k$

$x+y+z=12$

$\Rightarrow$$3k+4k+5k=12$

$\Rightarrow$$k=1$

$\therefore$ $x=3; y=\mathrm{4;} z=5$

Therefore, $x^3+y^3+z^3=216$.