It seems so easy

Algebra Level 4

if xa=yb=zc\frac { x }{ a } =\frac { y }{ b } =\frac { z }{ c } then [a2x2+b2y2+c2z2a3x+b3y+c3z]32\quad { \left[ \frac { { a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+{ c }^{ 2 }{ z }^{ 2 } }{ { a }^{ 3 }x+{ b }^{ 3 }y+{ c }^{ 3 }z } \right] }^{ \frac { 3 }{ 2 } } =

(the problem is not original)

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