Solution
= Solution
The square is even, and direct <integration> gives
$$
\int_{-1}^1f(x)^2\,dx
=2\int_0^1x^2(1-x)^2\,dx
=2\left(\frac13-\frac12+\frac15\right)=\frac1{15}.
$$
Using the nonzero <Fourier coefficients> in <Parseval's identity>,
$$
\frac1{30}=\frac12\sum_{m=0}^\infty\frac{64}{\pi^6(2m+1)^6}.
$$
Hence
$$
\boxed{\sum_{m=0}^\infty\frac1{(2m+1)^6}=\frac{\pi^6}{960},\qquad A=\frac1{960}}.
$$