If x + y + z = 9, then the value of (x - 4)³ + (y - 2)³ + (z - 3)³ - 3(x - 4)(y - 2)(z - 3) is?

Solution (By Examveda Team)

$$\eqalign{ & \mathop {\mathop {{{\left( {x - 4} \right)}^3}}\limits_ \Downarrow }\limits_{{a^3}} + \mathop {\mathop {{{\left( {y - 2} \right)}^3}}\limits_ \Downarrow }\limits_{{b^3}} + \mathop {\mathop {{{\left( {z - 3} \right)}^3}}\limits_ \Downarrow }\limits_{{c^3} - 3abc} - 3\left( {x - 4} \right)\left( {y - 2} \right)\left( {z - 3} \right) \cr & \Rightarrow a + b + c = x - 4 + y - 2 + z - 3 \cr & \Rightarrow a + b + c = x + y + z - 9 \cr & \Rightarrow a + b + c = 9 - 9 \cr & \Rightarrow a + b + c = 0 \cr} $$
$${\text{So, }}{\left( {x - 4} \right)^3} + {\left( {y - 2} \right)^3} + $$     $${\left( {z - 3} \right)^3} - $$   $$3\left( {x - 4} \right)$$  $$\left( {y - 2} \right)$$ $$\left( {z - 3} \right)$$
$$ \Rightarrow 0$$