In $\triangle ABC$, I is the incenter. Area of $\triangle IBC = 28$, area of $\triangle ICA= 30$ and area of $\triangle IAB = 26$. Find $AC^2 − AB^2$

we note $AC=b,BC=a,AB=c$, we will find $b^2-c^2$. We have $$ar=56,br=60,cr=52.$$ then we have the proportion of $a$ and $b,c$: $$b=\frac{60}{56}a,c=\frac{52}{56}a$$ Then we have the equality: $$a=\frac{56}{112}(b+c)=\frac{1}{2}(b+c)$$ $$a=\frac{56}{8}(b-c)=7(b-c)$$ Now we should recall the formula of triangle's area: $S=\sqrt{p(p-a)(p-b)(p-c)}$, where $p=\frac{a+b+c}{2}$ Then we have $$84=\sqrt{\frac{a+b+c}{2}\cdot\frac{b+c-a}{2}\cdot\frac{a+c-b}{2}\cdot\frac{a+b-c}{2}}\\=\frac{1}{4}\sqrt{[(b+c)^2-a^2][a^2-(b-c)^2]}\\=\frac{1}{4}\sqrt{[(b+c)^2-\frac{1}{4}(b+c)^2][49(b-c)^2-(b-c)^2]}\\=\frac{1}{4}\sqrt{\frac{3}{4}(b+c)^2\cdot48(b-c)^2}\\\frac{3}{2}|b^2-c^2|\\=\frac{3}{2}(b^2-c^2)$$(because $b>c$).

then we have what we want.

A remark on the solvability of the question: you are given $ra=56,rb=60,rc=52$ (with the usual notations $r$ the inradius, $a=BC$, etc). Therefore you can deduce that the triangle $ABC$ is similar to the triangle $\Delta=(56,60,52)$ with scaling ratio $r$.

You find the area of $ABC$ by summing up the areas of the three triangles and use the fact that $r^2 Area(ABC)=Area(\Delta)$ (you can calculate $Area(\Delta)$). This gives you $r$, and therefore you can find $a,b,c$.