Integrate By Parts Or Use The Beta Function

I'll use $\color{magenta}{\text{Integration by Parts} }$ $I_{m,n}=\int_0^1 (1-x^m)^n=[x(1-x^m)^n]^1_0 +mn \int_0^1 x^m(1-x^m)^{n-1}$ $=0-mn\int_0^1 (1-x^m-1)(1-x^m)^{n-1}$ $=-mn ( \int_0^1 x^m(1-x^m)^n- \int_0^1 x^m(1-x^m)^{n-1})$ $=mn(I_{m,n-1}-I_{m,n})$ $\Rightarrow I_{m,n}=mn(I_{m,n-1}-I_{m,n})$ $\Rightarrow \dfrac{I_{m,n}}{I_{m,n-1}}=\dfrac{mn}{mn+1}$ $\text{We are required to find} \quad \color{limegreen}{2mn+1} \quad\text{when m=n=1729}$ Hence required answer=2(1729)(1729)+1= $\Large 5978883$

This can be done by IBP or beta function .As i recently learned about beta fn i will post solution using it.

Let $I_m={\int_0^1 (1-x^{c})^{m}dx}$

Let $x^c=t$

$cx^{c-1}dx=dt$

also $x^{c-1}=t^{\frac{c-1}{c}}$

So $I_m=\dfrac{1}{c}{\int_0^1 (1-t)^{m}(\dfrac{1}{t^{\frac{c-1}{c}}})dt} \Rightarrow\dfrac{1}{c}{\int_0^1 (1-t)^{m}(t^{\frac{1}{c}-1})dt}$

Hence $I_m=\dfrac{1}{c} \beta (m+1,\dfrac{1}{c})$

Similarly $I_{m-1}=\dfrac{1}{c} \beta (m,\dfrac{1}{c})$

$\dfrac{I_{m}}{I_{m-1}}=\dfrac{ \beta (m+1,\dfrac{1}{c})}{ \beta (m,\dfrac{1}{c})}$

Using $\beta (x,y)= \dfrac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$

$\dfrac{I_{m}}{I_{m-1}}=\dfrac{\Gamma(m+\frac{1}{c}) \Gamma(m+1)}{\Gamma(m+\frac{1}{c}+1) \Gamma(m)}$

Using $\Gamma(s+1)=s \Gamma(s)$

$\dfrac{I_{m}}{I_{m-1}}=\dfrac{cm}{cm+1}$

Put $c=m=1729$ and get answer.