Without actually computing the cubes, find the value of : \( 48^3 – 30^3 – 18^3 \)

\[ 48^3 – 30^3 – 18^3 \] \[ = 48^3 + (-30)^3 + (-18)^3 \] \[ \text{Since } 48 + (-30) + (-18) = 0 \] \[ \therefore a^3 + b^3 + c^3 = 3abc \quad \text{when } a+b+c=0 \] \[ = 3(48)(-30)(-18) \] \[ = 77760 \]