Three 3's

Express $x^3(1-x^3)^3$ as $[1-(1-x^3)](1-x^3)^3$ = $(1-x^3)$ - $(1-x^3)^4$ . The integral of $(1-x^3)$ from 0 to 1 with respect to x is represented by A as given in the question.

To integrate $(1-x^3)^4$ with respect to x, we have $\frac{du}{dx}$ = 1 and v= $(1-x^3)^4$ so that u = x and $\frac{dv}{dx}$ = $4(-3x^2)(1-x^3)^3$ = $-12x^2(1-x^3)^3$ . Now we have $x(1-x^3)^4$ - the integral of $-12x^2(1-x^3)^3$ or $x(1-x^3)^4$ - the integral of $x(-12x^2)(1-x^3)^3$ or $x(1-x^3)^4$ + the integral of $12x^3(1-x^3)^3$

You would realise that upon substitution of 0 and 1, $x(1-x^3)^4$ would be reduced to 0 so we have the integral of $(1-x^3)$ - $(1-x^3)^4$ as A - the integral of $(12x^3)(1-x^3)^3$ or A - 12 times the integral of $x^3(1-x^3)^3$ This is also the integral of $x^3(1-x^3)^3$ .

Hence, A is 12+1 = 13 times the integral of $x^3(1-x^3)^3$ .

You can use Beta function and then multiply the first integral, A, by the inverse of the second, k/A, to obtain K. Knowing that gamma(x+1)=x*gamma(x), the problem is easy. K = 13