Step 1: Count the carbons.
The chain $CH_3-C\equiv C-CH=CH-CH=CH-CH_3$ has $8$ carbons, so the root is oct.
Step 2: List the multiple bonds.
There is one triple bond and two double bonds, so the suffix combines diene and yne, giving an oct...dien...yne name.
Step 3: Decide numbering direction.
Number from the end that gives the lowest locants to the multiple bonds as a set. From the left, the triple bond starts at C2 and the double bonds at C4 and C6. From the right they would be higher, so number from the left.
Step 4: Apply the tie-break rule.
Both ends could be argued, but the rule says when there is a choice the double bonds get the lower numbers over the triple bond. Numbering from the left gives double bonds the higher numbers here, so we instead check the set $\{2,4,6\}$ which is lowest overall and keeps the triple bond at $2$.
Step 5: Build the name.
Triple bond at $2$, double bonds at $4$ and $6$: the name is octa-... With en written before yne, it reads octa-2-... Wait, the correct lowest-locant set places it as octa-2,4-dien-6-yne.
Step 6: Final name.
Hence the IUPAC name is octa-2,4-dien-6-yne.
\[ \boxed{\text{Octa-2,4-dien-6-yne}} \]