[Terrytw]

Here is what I have done so far. Can anyone offer suggestions?

1^3 + 2^3 + 3^3 + * * * + n^3 = 1/4(n^2)[(n+1)^2]

Let P(1) = 1 then 1^3 = 1/4(1^2)[(1+1)^2]=1 so P(1) is true.

So we assume since it is true for P(1) then it must be true for P(k) as well and
then need to prove true for P(k+1)

so P(k) = 1^3 + 2^3 + 3^3 + * * * + k^3 = 1/4(k^2)[(k+1)^2] = 1/4[k(k+1)]^2

P(k+1) = 1^3 + 2^3 + 3^3 + * * * + k^3 + (k^3+1)= 1/4(k^2)[(k+1)^2] = 1/4[(k+1)(k+1)+1]^2

and this is where I get stuck.

[ebaines]

You want to show that P(k+1) = P(k) + (k+1)^3. You wrote (k^3+1), and that's not correct.

If you substitute (k+1) in place of k in the formula P(k) = 1/4[k^2(k+1)^2], you can expand it out, and ultimately show that:

P(k+1) = P(k) + (k+1)^3

Write back if you're still having difficulty.