X still unknown

Let

$\begin{aligned}x^4+x^2y^2+y^4=&\ 24\Rightarrow(1)\\x^2-xy+y^2=&\ 6\Rightarrow(2)\end{aligned}$

From Equation $(1)$ That is,

$\begin{aligned}x^4+x^2y^2+y^4=&\ 24\\(x^4+2x^2y^2+y^4)-x^2y^2=&\ 24\\(x^2+y^2)^2-(xy)^2=&\ 24\\(x^2+y^2-xy)(x^2+y^2+xy)=&\ 24\\6(x^2+xy+y^2)=&\ 24\\x^2+xy+y^2=&\ 4\Rightarrow(3)\end{aligned}$

Equation $(2)$ $+$ Equation $(3)$ That is,

$\begin{aligned}2x^2+2y^2=&\ 10\\x^2+y^2=&\ 5\Rightarrow(4)\end{aligned}$

Equation $(2)$ $-$ Equation $(3)$ That is,

$\begin{aligned}-2xy=&\ 2\Rightarrow(5)\end{aligned}$

Equation $(4)$ $-$ Equation $(5)$ That is,

$\begin{aligned}x^2+2xy+y^2=&\ 3\\(x+y)^2=&\ 3\\x+y=&\ \pm\sqrt3\end{aligned}$

From $|x^3+y^3|=|(x+y)(x^2-xy+y^2)|$

When $x+y=\sqrt3$ the value of $|x^3+y^3|$ is $|\sqrt3\cdot6|=6\sqrt3$

When $x+y=-\sqrt3$ the value of $|x^3+y^3|$ is $|-\sqrt3\cdot6|=6\sqrt3$

Hence, the value of $|x^3+y^3|$ is $\boxed{6\sqrt3}$

Simply another approach in detail. Same as Ambuj Kumar Pandit :-
$(x^2-xy+y^2)^2 -(x^4+x^2y^2+y^4)=36-24=12.\\ \implies~-xy(x^2+y^2)+x^2y^2=6..\\ \therefore~xy= - \dfrac 6{x^2-xy+y^2}=- \dfrac 6 6 =-1\\ \therefore~x+y={\large \sqrt{x^2-xy+y^2+3xy} }=\sqrt{6-3}=\sqrt3\\ |x^3+y^3|=(x+y)(x^2-xy+y^2)=\sqrt3 * 6$

Using the Sophie-Germain Identities: $\ x^4 +x^2 y^2 +y^4 = (x^2 +xy+y^2)(x^2 -xy+y^2) \Rightarrow\ 24 = 6(6+2xy)$ $\therefore\ xy = -1$ Now using the 2nd given equation: $\ x^2 +y^2 = 6+xy = 5 \Rightarrow\ (x+y)^2 = 3$ $\therefore\ x+y = \sqrt{3}$ Using the factor theorem we know that $\ x+y$ is a factor of $\ x^3 +y^3$ which produces: $\ |x^3 +y^3| = |(x+y)(x^2 -xy +y^2)| = \boxed{+6 \sqrt{3} }$