已知ax³=by³=cz³,且1/x+1/y+1/z=1 試證(ax²+by²+cz²)^1/3=a^1/3+b^1/3+c^1/3?

ax³ = by³ = cz³ ⇒ ∛ax = ∛by = ∛cz
令 ∛ax = ∛by = ∛cz = k ,
則 ∛(ax² + by² + cz²)
= ∛((k/x)³ x² + (k/y)³ y² + (k/z)³ z²)
= ∛(k³/x + k³/y + k³/z)
= ∛(k³(1/x + 1/y + 1/z))
= ∛k³
= k
= k(1/x + 1/y + 1/z)
= k/x + k/y + k/z
= ∛a + ∛b + ∛c