You reverse what you did, so first you go 5 units down which makes the coordinates (9,1) then you move 3 units left which then is (6,1). Then you go 4 units up which lets you end out at (6,5).
Start by decomposing the number inside the root into primes
Then group the terms into cubes if possible

rewrite the root
![\sqrt[3]{80}=\sqrt[3]{10\cdot2^3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B80%7D%3D%5Csqrt%5B3%5D%7B10%5Ccdot2%5E3%7D)
then cancel the terms that are cubes and bring them out of the root
487 rounded to the nearest ten is 490......because 487 is closer to 490 then it is to 480.