But I can seem to get this part:
Assuming P(k) is true, you can cancel it from the left hand side if you also cancel the right side of P(k) from the right side of P(k+1). You should be left with an identity, which proves that P(k) implies P(k+1).

OK, note that the part in
blue below is just P(k).
Σ_{j=1}^{k}j^{3}+(k+1)
^{3}=(1/4)(k+1)
^{2}(k+2)
^{2}
What you need to do is expand the
right hand side of both P(k) and of P(k+1). Then, you can subtract Σ
_{j=1}^{k}j
^{3} from the left side of the above, and you can also subtract (1/4)x
^{2}(x+1)
^{2} from the right side of the above (since those two quantities are
equal).
What you will be left with is:
(k+1)
^{3}=(1/4)(4k
^{3}+12k
^{2}+12k+4)
or...
(k+1)
^{3}=k
^{3}+3k
^{2}+3k+1,
which is an identity.
edit: fixed an omission