Balance by oxidation number metod :
`CU+HNO_(3)toCu(NO_(3))_(2)+NO+H_(2)O`

Explanation

In this reaction Cu undergoes ,while `HNO_(3)` undergoes reduction. The oxidation number of Cu increases by 2 unit (0 to +2) and that of N decreases by 3 units (+5 to +2). Therefore ,two `HNO_(3)` molecules
for three cu-atoms are needed to balance the increases and decrease in oxidation numbers .Hence ,the equation will be
`3Cu+2HNO_(3)to3Cu(NO_(3))_(2)+2NO+4H_(2)O`
Now ,6 molecules of `HNO_(3)`are needed for the formation of 3 molecules of `Cu(NO_(3))_(2)` .So additional 6 molecule of `HNO_(3)` are to be added on the left side of the above equation .Hence ,the balanced equation for the given reaction will be-
`3Cu+2HNO_(3)to3Cu(NO_(3))_(2)+2NO+4H_(2)O`