It's not clear what 5i3/n8=(5/n8)(i3) means, so I'll assume that you're referring to the sum of a series:
(5/n8)∑i3 where integer i: 1≤i≤n.
That is Sn=(5/n8)(1+8+27+...+(n-1)3+n3)=(5/n8)(n(n+1)/2)2.
Sn=5(n+1)2/(4n6).
INDUCTIVE PROOF
If sum of cubes of the natural numbers is Sn=(n(n+1)/2)2, then:
Sn+1=Sn+(n+1)3. S1=1, and S2=9=(2×3/2)2, so base case is true.
Sn+1 should be ((n+1)(n+2)/2)2=(n2+3n+2)/4=(n4+6n3+13n2+12n+4)/4.
Sn+(n+1)3=(n4+2n3+n2)/4+n3+3n2+3n+1=
(n4+2n3+n2+4n3+12n2+12n+4)/4=(n4+6n3+13n2+12n+4)/4=Sn+1. This proves by induction that the formula used is correct.