Determine the value of c that makes the blue area above y = c equal to the blue area below y = c. 
Determine the value of c that makes the blue area above y = c equal to the blue area below y = c.
edit: I'm kind of stuck on this problem, not sure what steps to do so that I can find the equal areas.
edit2: The answer looks right so far, thanks to everyone who helped out!
 A: Some things to point out:
We take the integral of a function $f(x)$ to find the area under the curve. What we are essentially doing is finding the area $A=\int_a^b f(x)-0\quad\!\!\!dx$ which is the area under the function but above the $x$-axis.
So, the area of the left region would be $\int_0^a c-(8x-27x^3)dx$ and the area of the right region would be $\int_a^b 8x-27x^3-c\quad\!\!\!dx$, where $a$ and $b$ are the intersections of the two functions (when $8x-27x^3=c$). 
But these areas have to be equal so the integrals so $\int_0^a c-(8x-27x^3)dx$=$\int_a^b 8x-27x^3-c\quad\!\!\!dx$
A: Let $x=a$ be the first intersection and $x=b$ be the second.
The areas are equal, so $\displaystyle\int_0^a c-(8x-27x^3)dx=\int_a^b 8x-27x^3-cdx$. Carrying out the integration, we have:
$$cx-4x^2+\frac{27}{4}x^4\Bigg|_0^a= 4x^2-\frac{27}{4}x^4-cx\Bigg|_a^b$$
$$ca-4a^2+\frac{27}{4}a^4= 4b^2-\frac{27}{4}b^4-cb-4a^2+\frac{27}{4}a^4+ca$$
$$0= 4b^2-\frac{27}{4}b^4-cb=b(4b-\frac{27}{4}b^3-c)$$
Thus, $c=4b-\dfrac{27}{4}b^3$
Now, we also recall that $f(b)=8b-27b^3=c$
We set the two equations equal giving: 
$$4b-\dfrac{27}{4}b^3=8b-27b^3$$
$$0=4b-\dfrac{81}{4}b^3$$
By normal methods, we find that $b=\dfrac{4}{9}$. We substitute this back into our equation relating $b$ and $c$ to find that: $$f(\dfrac{4}{9})=8\dfrac{4}{9}-27(\dfrac{4}{9})^3=\dfrac{32}{27}=c$$
