Radical Equation 1

$\sqrt{10 + \sqrt{x^3 + 100}} = 10 - \sqrt{x^3 + 100}$

$a = \sqrt{x^3 + 100}$

$\sqrt{10 + a} = 10 - a$

$10 + a = 100 - 20a + a^2$

$a^2 - 21a + 90 = 0$

$(a - 15)(a - 6) = 0$

$a = 15,~ 6$

$\sqrt{x^3 + 100} = 15,~ 6$

$x^3 + 100 = 225,~ 36$

$x^3 = 125,~ -64$

$x = 5, -4$

$\text{If }x = 5,~ \sqrt{10 + \sqrt{5^3 + 100}} = 10 - \sqrt{5^3 + 100}$

$\sqrt{10 + \sqrt{225}} = 10 - \sqrt{225}$

$\sqrt{10 + 15} = 10 - 15$

$\sqrt{25} = -5$

$\sqrt{x} = \mid\sqrt{x}\mid \therefore \sqrt{25} = 5$

$5 \neq -5 \therefore x \neq 5,~ x = -4$

$\large \boxed{x = -4}$

Let $\sqrt{10 + \sqrt{x^3 + 100}}=\alpha \Rightarrow \sqrt{x^3 + 100}=\alpha^2-10$ . Equation transforms to: $\alpha=-\alpha^2+20$ $\Rightarrow \alpha^2+\alpha-20=0$ $\alpha=4=\sqrt{10 + \sqrt{x^3 + 100}}$ (-5 rejected since $\alpha>0$ ) $\Rightarrow x^3=-64\Rightarrow x=\huge\boxed{\color{#007fff}{-4}}$