Induction - have answer, need clarification

[tangibleLime]

Homework Statement

Example 3 from this website:
http://www.themathpage.com/aprecalc/mathematical-induction.htm

Prove this formula for the sum of consecutive cubes:
1³ + 2³ + 3³ + . . . + n³ = (n²(n + 1)²)/4

Homework Equations

The Attempt at a Solution

I can understand and follow the entire thing up to a certain point. It seems like the algebra they start to do just doesn't follow. Where I get lost is at the location where they " To do that, add the next cube to S(k)". On the third step, they go from (k²(k + 1)² + 4(k + 1)³)/4 to ((k + 1)²[k² + 4(k + 1)])/4, which I am certain is nonsense. To back up my claims, I threw it into WolframAlpha and it said that those two expressions are not equal.

When I tried to do this myself, I just calculated (k+1)^3 and expanded it to (k^3+3k^2+3k+1), which is completely different from they they did. It's as if they just dropped the cube on the (k+1)?

Any clarification would be great.

 

[Dick]

On the third step, they go from (k²(k + 1)² + 4(k + 1)³)/4 to ((k + 1)²[k² + 4(k + 1)])/4, which I am certain is nonsense. To back up my claims, I threw it into WolframAlpha and it said that those two expressions are not equal.

When I tried to do this myself, I just calculated (k+1)^3 and expanded it to (k^3+3k^2+3k+1), which is completely different from they they did. It's as if they just dropped the cube on the (k+1)?

Any clarification would be great.

Um, I actually think that is true. You might have put it into Wolfram Alpha wrong. You can expand the cube, sure, but then you'll just want to factor it again. They just saved some time by taking advantage of common factors and not expanding.

 

[tangibleLime]

Ooooh, okay, thanks! Just worked out the whole thing by hand and seems like it flows nicely now. Maybe I made a small mistake before.

Thanks again!

 

[Mentallic]

Like Dick said, they factorized out using common factors.

Say you wanted to find the factored form of $$(a+b)^3+2a(a+b)^2$$
rather than expanding everything out and then factorizing from there, which can be tedious and difficult to solve, they noticed that there is a common factor of $$(a+b)^2$$ and thus can factorize it as so: $$(a+b)^2\left((a+b)^1+2a\right)=(a+b)^2(3a+b)$$.