Unknown sides, known area

Let the ratio within the segments be $k$ .

Now, we can see that $A_{\Delta{AYC}} = kA_{\Delta{ABC}}$ , as if we were to treat this side as the base, with the height unchanged, a shift in the base length with some ratio will also result to the same shift in area.

In a similar manner, we can see that $A_{\Delta{AYX}} = kA_{\Delta{AYC}}$ , which in turn implies that $A_{\Delta{AYX}} = k^2 A_{\Delta{ABC}}$ . Then, once again, we will find that $A_{\Delta{AZX}} = kA_{\Delta{AYX}}$ , implying finally that $A_{\Delta{AZX}} = k^3 A_{\Delta{ABC}}$ . We will use this later.

We can also start at $\Delta{BXC}$ , proceed to $\Delta{BXY}$ , then to $\Delta{BZY}$ . We will eventually arrive at $A_{\Delta{BZY}} = (1-k)^3 A_{\Delta{ABC}}$ .

Now, observe that

$( A_{\Delta{ZAX}})^{\frac{1}{3}} = k (A_{\Delta{ABC}})^{\frac{1}{3}}$

$(A_{\Delta{BZY}})^{\frac{1}{3}} = (1- k)(A_{\Delta{ABC}})^{\frac{1}{3}}$

adding them all, we get

$A_{\Delta{ZAX}})^{\frac{1}{3}} + (A_{\Delta{BZY}})^{\frac{1}{3}} = (A_{\Delta{ABC}})^{\frac{1}{3}}$

and cubing both sides, we arrive at

$(A_{\Delta{ZAX}})^{\frac{1}{3}} + (A_{\Delta{BZY}})^{\frac{1}{3}})^{3} = (A_{\Delta{ABC}})$

and so we can now find the area of $\Delta ABC$ , and that is

$( 8^{\frac{1}{3}} + 27^{\frac{1}{3}})^3$

$(2+3)^3$

$5^3 = \boxed{125}$