Please demonstrate the underlying algebra between these expressions?

[HRN]

I'm having trouble working through a proof via induction to a problem in Serge Lang's Basic Mathematics; not because of the theory, but because I don't understand the algebra.

Simply put, if someone would demonstrate every step in the algebra between the following two expressions, it would aid my development in maths (and I'd be very grateful):

= ((n (n+1)) /2)² + (n+1)³
= (1/4) (n+1)² (n²+4n+4)

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Here's the full problem for reference (with the part I don't get emboldened), as well as the solution in the answer key.

Prove that, for all integers n >= 1, we have
1³ + 2³ + ... + n³ = ((n (n+1)) /2)².

Proof.
Let A(n) = 1³ + 2³ + ... + n³ (written in sigma notation with the index of summation k³, lower bound 1 and upper bound n). We want to prove that A(n) = ((n (n+1)) /2)².
For A(1) we have 1 = (1 (2) /2)² = 1. By induction,
A(n+1) = A(n) + (n+1)³ = (1/4) (n+1)² (n²+4n+4).
Thus, A(n+1) = (1/4) (n+1)² (n+2)² = (((n+1)² (n+2)²) /2)².

Sorry in advance if that appears too messy; this is my first post here!

 

[soroban]

Hello, HRN!

\(\displaystyle \left[\dfrac{n(n+1)}{2}\right]^2 + (n+1)^3 \;\;\rightarrow\;\;\frac{1}{4}(n+1)^2(n^2+4n+4)\)

We have: .\(\displaystyle \dfrac{n^2(n+1)^2}{4} + (n+1)^3 \;=\;\dfrac{n^2(n+1)^2 + 4(n+1)^3}{4}\)

Factor: .\(\displaystyle \dfrac{(n+1)^2\big[n^2 + 4(n+1)\big]}{4} \;=\;\dfrac{(n+1)^2(n^2 + 4n + 4)}{4} \)

. . . . . . \(\displaystyle =\;\dfrac{(n+1)^2(n+2)^2}{4} \;=\;\left[\dfrac{(n+1)(n+2)}{2}\right]^2\)

 

[HRN]

Thanks soroban, I understand it now. It's hard to visualize; I'll have to study this more.
Very helpful!