# 300+ followers problem

Algebra Level 5

$\large \dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{x-z}$

Let $$m$$ be the minimum value of the above expression for reals $$x > y > z$$ given $$(x − y)(y − z)(x − z) = 300$$.

Given that $$m$$ can be expressed as $$\dfrac{1}{A}\sqrt[3]{\dfrac{B}{C}}$$ where $$A, B, C$$ are positive integers and $$\gcd(B,C)=1$$, with $$B+C$$ minimized.

Find $$A+B+C$$.

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