If $$x = \root 3 \of {a + \sqrt {{a^2} + {b^3}} } $$     + $$\root 3 \of {a - \sqrt {{a^2} + {b^3}} } $$     then $${x^3} + 3bx$$   is equal to?

Solution(By Examveda Team)

 $${\text{Given}}\,x = \root 3 \of {a + \sqrt {{a^2} + {b^3}} } \,\,+ $$      $$\root 3 \of {a - \sqrt {{a^2} + {b^3}} } $$
$$\eqalign{ & {\text{Let}}\,z = \sqrt {{a^2} + {b^3}} \cr & \therefore x = \root 3 \of {a + z} + \root 3 \of {a - z} \cr & {\text{Cubing both side}} \cr} $$
  $$\therefore {x^3} = {\left( {\root 3 \of {a + z} } \right)^3} + {\left( {\root 3 \of {a - z} } \right)^3} \, + $$      $$\,3{\left( {a + z} \right)^{\frac{1}{3}}}$$  $${\left( {a - z} \right)^{\frac{1}{3}}} \times\, $$   $$\left( {\root 3 \of {a + z} + \root 3 \of {a - z} } \right)$$
 $$ \,\Rightarrow {x^3} = a + z + a - z \,+\, $$    $$3{\left( {{a^2} + az - az - {z^2}} \right)^{\frac{1}{3}}}$$    $$ \times \left( x \right)$$
$$\eqalign{ & \Rightarrow {x^3} = 2a + 3{\left( {{a^2} - {z^2}} \right)^{\frac{1}{3}}} \times \left( x \right) \cr & {\text{Put the value of }}z \cr & \Rightarrow {x^3} = 2a + 3{\left( {{a^2} - {a^2} - {b^3}} \right)^{\frac{1}{3}}} \times \left( x \right) \cr & \Rightarrow {x^3} = 2a + 3{\left( { - {b^3}} \right)^{\frac{1}{3}}} \times \left( x \right) \cr & \Rightarrow {x^3} = 2a - 3bx \cr & \therefore {x^3} + 3bx = 2a \cr} $$