Best to plot the curves and lines on the same graph so that the region is plain to see, especially its boundaries.
Relevant intersection point forming the boundaries of the enclosed region:
Leftmost limit is solution of x²-3x+2=2x+8, x²-5x-6=0=(x+1)(x-6).
So x=-1 is the leftmost limit. Intersection point (-1, 6).
Rightmost limit is solution of x³-1=15-3x, x³+3x-16=0.
This has the approximate solution x=2.12677. x=2.1 to 1 decimal place. Intersection point (2.1, 8.6).
Highest point is when 2x+8=15-3x, 5x=7, x=7/5 or 1.4. Intersection point is (1.4, 10.8).
Lowest point is when x³-1=x²-3x+2, (x-1)(x²+3)=0.
The only solution is x=1. Intersection point is (1,0).
The question isn’t clear about whether the area of the region is also required.
The region is subdivided into 3 areas with x intervals as shown:
A₁ [-1,1] ∫(2x+8-x²+3x-2)dx=∫(6+5x-x²)dx=
[6x+5x²/2-x³/3]¹₋₁=12-⅔=34/3.
A₂ [1,1.4] ∫(2x+8-x³+1)dx=∫(9+2x-x³)dx=
[9x+x²-x⁴/4]¹˙⁴₁=3.8496.
A₃ [1.4,2.12677] ∫(16-3x-x³)dx=[16x-3x²/2-x⁴/4]²˙¹²⁶⁷⁷₁.₄=3.62927.
Total area=34/3+3.8496+3.62927=18.8 to 1 decimal place.