1. **Problem statement:** We are given a positive integer $n$ with exactly 8 positive divisors, including 1 and $n$. Two of these divisors are 14 and 21. We need to find the sum of all 8 positive divisors of $n$. 2. **Key idea:** The number of divisors function depends on the prime factorization of $n$. If $$n = p^a q^b r^c \cdots,$$ then the number of positive divisors is $$(a+1)(b+1)(c+1) \cdots = 8.$$ 3. **Analyze given divisors:** Since 14 and 21 divide $n$, their prime factors must be in $n$: - $14 = 2 \times 7$ - $21 = 3 \times 7$ So $n$ must be divisible by primes 2, 3, and 7. 4. **Form of $n$:** Let $$n = 2^a \times 3^b \times 7^c,$$ where $a,b,c \geq 1$ because 14 and 21 divide $n$. 5. **Number of divisors:** $$(a+1)(b+1)(c+1) = 8.$$ Since 8 factors as $8=8\times1\times1=4\times2\times1=2\times2\times2$, possible triples for $(a+1,b+1,c+1)$ are permutations of these. 6. **Check possible triples:** - $(4,2,1)$ means $a=3,b=1,c=0$ but $c=0$ contradicts $7|n$. - $(2,2,2)$ means $a=b=c=1$ which fits all conditions. So $a=b=c=1$. 7. **Calculate $n$:** $$n = 2^1 \times 3^1 \times 7^1 = 42.$$ 8. **Find all divisors of 42:** Divisors are all numbers of the form $2^i 3^j 7^k$ where $i,j,k \in \{0,1\}$: $$1, 2, 3, 6, 7, 14, 21, 42.$$ 9. **Sum of divisors:** $$1 + 2 + 3 + 6 + 7 + 14 + 21 + 42 = 96.$$ **Final answer:** $$\boxed{96}.$$