1. State the problem: Simplify the expression $w=\frac{\sqrt{8^{-6}}\times3^4\times\sqrt{10^{12}}}{5^6\times\sqrt{6^8}}$. 2. Key formulas and rules used: Use $\sqrt{a}=a^{1/2}$ and therefore $\sqrt{a^b}=a^{b/2}$. 3. Additional exponent rules: Use $(ab)^c=a^c b^c$, $a^{m}a^{n}=a^{m+n}$, and $(a^m)^n=a^{mn}$. 4. Convert each radical to an exponent: $\sqrt{8^{-6}}=(8^{-6})^{1/2}=8^{-3}$. 5. Convert the others similarly: $\sqrt{10^{12}}=(10^{12})^{1/2}=10^6$. 6. And: $\sqrt{6^8}=(6^8)^{1/2}=6^4$. 7. Substitute these into $w$: $w=\frac{8^{-3}\times3^4\times10^6}{5^6\times6^4}$. 8. Factor composite bases into primes: $8=2^3$ so $8^{-3}=2^{-9}$. 9. Also $10=2\cdot5$ so $10^6=2^6\cdot5^6$. 10. And $6=2\cdot3$ so $6^4=2^4\cdot3^4$. 11. Substitute prime factorizations: $w=\frac{2^{-9}\times3^4\times2^6\times5^6}{5^6\times2^4\times3^4}$. 12. Combine powers of the same base: for base 2, $2^{-9}\times2^6=2^{-3}$. 13. Cancel identical factors $3^4$ and $5^6$ from numerator and denominator. 14. After cancellation we have $w=\frac{2^{-3}}{2^4}=2^{-7}$. 15. Convert negative exponent to fraction: $2^{-7}=\frac{1}{2^7}=\frac{1}{128}$. 16. Final answer: $w=\frac{1}{128}$.