Find the value of the following expression $$ \prod^{\infty}_{n=0}\left(1-\frac{1}{2^{3n}}-\frac{1}{2^{6n}}\right)$$
What I tried:
If $\frac{1}{2^3}=x$, then we can write expression as
$$\prod^{\infty}_{n=0}(1-x^n-x^{2n})=(1-1+1)(1-x-x^2)(1-x^2-x^4)(1-x^3-x^6)\cdots$$
How do I find that infinite product?
This was for the first version of the question. $$\prod^{\infty}_{n=0}\left(1-\frac{1}{2^{3n}}-\frac{1}{2^{6n}}\right)=\prod^{2}_{n=0}\left(1-\frac{1}{2^{3n}}-\frac{1}{2^{6n}}\right)\prod^{\infty}_{n=3}\left(1-\frac{1}{2^{3n}}-\frac{1}{2^{6n}}\right)$$
The first product is $-\frac{221705}{262144}$. For the second one, define $$a_n=\left(1-x^n-x^{2n}\right)\implies \log(a_n)=\log\left(1-x^n-x^{2n}\right)$$ Now, using Taylor $$\log(a_n)\sim -x^n-x^{2n}\implies \sum_{n=3}^\infty \log(a_n)\sim\frac{x^6+x^4+x^3}{(x-1) (x+1)}$$ Make $x=\frac 18$ to get $$\sum_{n=3}^\infty \log(a_n)\sim -\frac{577}{258048}\implies \prod^{\infty}_{n=3}\left(1-\frac{1}{2^{3n}}-\frac{1}{2^{6n}}\right)\sim \exp{\left(-\frac{577}{258048} \right)}$$ So, an estimate of the infinite product is $$-\frac{221705}{262144}\exp{\left(-\frac{577}{258048} \right)}\approx -0.8438484857$$ while the "exact" value is $-0.8438468422$
Edit
For the new version of the problem, the first product is $\frac{229881}{262144}$ $$\log(a_n)\sim -x^n+x^{2n}\implies \sum_{n=3}^\infty \log(a_n)\sim\frac{-x^6+x^4+x^3}{(x-1) (x+1)}$$ Make $x=\frac 18$ to get $$\sum_{n=3}^\infty \log(a_n)\sim -\frac{575}{258048}\implies \prod^{\infty}_{n=3}\left(1-\frac{1}{2^{3n}}+\frac{1}{2^{6n}}\right)\sim \exp{\left(-\frac{575}{258048} \right)}$$ and the estimate becomes $$\frac{229881}{262144}\exp{\left(-\frac{575}{258048} \right)}\approx 0.8749745708$$while the "exact" value is $0.8749728798$