More Radical Attack

Let $y=\sqrt[4]{x+8}$ . We can therefore write $\sqrt[3]{x-9}$ in terms of $y$ . To wit: $y=\sqrt[4]{x+8} \implies y^4=x+8 \implies y^4-17=x-9 \implies \sqrt[3]{x-9}=\sqrt[3] {y^4-17}$ It follows from the original equation that $\sqrt[3]{y^4-17}+y=7$ and notice that one of the radicals have been removed, which makes it easier for us to solve the equation. Thus $\sqrt[3]{y^4-17}=7-y \implies y^4-17=(7-y)^3=343-147y+21y^2-y^3 \implies y^4+y^3-21y^2+147y-360=0$ By the Rational Root Theorem, we obtain $y=3$ and the depressed equation will be $y^3+4y^2-9y+120=0$ . We can show that this equation has no rational root [although I will not show it here, because I am not too fluent typing mathematical equations using LaTeX]. Thus $y=3$ . And therefore from $y=\sqrt[4]{x+8}$ , $3=\sqrt[4]{x+8} \implies 81=x+8 \implies \boxed{x=73}$

easiest way is to plug in values of x such that (x-9) term is a perfect cube @ x-9 =64 x==73 thx... sometimes its easier to bruteforce ...

73 is the only possible number that will lead to the exact value of 7 for the given cubic and quartic value