If x = 3 a + a^2 + b^3 + 3 a - a^2 + b^3 then x^3 + 3bx is equal to?

Answer: Option C

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