a) Ta có:
$\frac{AB^3}{AC^3}=\frac{AB^2}{AC^2}\cdot\frac{AB}{AC}=\frac{BE^2}{CF^2}\cdot\frac{BE}{CF}=\frac{BE}{CF}$
Vì $BE=AH\cdot\sin\angle BAH$ và $CF=AH\cdot\sin\angle CAH$, nên ta có:
$\frac{BE}{CF}=\frac{AH\cdot\sin\angle BAH}{AH\cdot\sin\angle CAH}=\frac{\sin\angle BAH}{\sin\angle CAH}=\frac{BH}{CH}$
Vậy ta có $\frac{AB^3}{AC^3}=\frac{BE}{CF}=\frac{BH}{CH}$

b) Ta có:
$AH^3=AH^2\cdot AH=AH^2\cdot AE=AH^2\cdot AH\cdot\sin\angle BAH=AH\cdot BH\cdot\sin\angle BAH=BC\cdot HE\cdot\sin\angle BAH=BC\cdot HE\cdot\sin\angle BAC$
Vì $\sin\angle BAC=\sin\angle CAH=\frac{CF}{AH}$, nên ta có:
$AH^3=BC\cdot HE\cdot\sin\angle BAC=BC\cdot HE\cdot\frac{CF}{AH}=BC\cdot HE\cdot\frac{CF}{AH}$
Vậy ta có $AH^3=BC\cdot HE\cdot CF$

c) Ta có:
$AH^3=BC\cdot HE\cdot CF=BC\cdot BE\cdot CF$
Vậy ta có $AH^3=BC\cdot BE\cdot CF$