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$\Rightarrow x^3+\dfrac{1}{x^3}=18$

$=(x+\dfrac{1}{x})(x^2+\dfrac{1}{x^2}-1)=18$

$=(x+\dfrac{1}{x})[(x+\dfrac{1}{x})^2-3]=18$

Let $(x+\dfrac{1}{x})=a$

$=(a)(a^2-3)=18$
$=(a)(a^2-3)=3×6$
$\therefore a=3$

$\Rightarrow x^4+\dfrac{1}{x^4}$

$=[(x+\dfrac{1}{x})^2-2]^2-2$

$=[(3)^2-2]^2-2$

$=(7)^2-2$

$\Rightarrow \boxed{47}$

A straightforward solution: $x^{3}+\frac {1}{x^{3}}=18$ . We get a common denominator: $\frac {x^{9}}{x^{3}}+\frac {1}{x^{3}}=18$ . We can rewrite the equation like so: $x^{6}-18x^{3}-1=0$ . We can solve using u-substitution: $u^{2}-18u-1=0$ . We get a solution of $u\approx18.06$ . And $18.06^{\frac{1}{3}}\approx 2.6$ . So to solve, $2.6^{4}+\frac {1}{2.6^{4}}\approx47$ . So the answer is $\boxed {47}$ .