(a)
The difference between consecutive cubes can be expressed as:
(i^3 – (i-1)^3) = 3i^2 – 3i + 1

To find the sum of the squares from 1 to n, we can sum both sides of the equation from i = 1 to i = n:
Σ(i^2) = Σ(i^3) – Σ((i-1)^3)

On the right side of the equation, we can substitute the difference between consecutive cubes:
Σ(i^2) = Σ(3i^2 – 3i + 1)

Now we can evaluate the summation by substituting the value of the lower and upper limits:
Σ(i^2) = (3(1^2) – 3(1) + 1) + (3(2^2) – 3(2) + 1) + … + (3(n^2) – 3(n) + 1)

Simplifying the above expression, we get:
Σ(i^2) = 3(1 + 4 + 9 + … + n^2) – 3(1 + 2 + 3 + … + n) + n

Using the formula for the sum of squares, the first term on the right side evaluates to:
1^2 + 2^2 + … + n^2 = n(n+1)(2n+1)/6

And, by the formula for the sum of natural numbers, the second term on the right side evaluates to:
1 + 2 + … + n = n(n+1)/2

So, substituting these values back into our expression for Σ(i^2), we get:
Σ(i^2) = 3(n(n+1)(2n+1)/6) – 3(n(n+1)/2) + n

and combining like terms
Σ(i^2) = n(n+1)(n+2)/3

So, we have derived the formula for the sum of the squares from 1 to n.

(b)
To find a formula for Σ(i^3), we can use a similar argument as before. We can start by finding the difference between consecutive fourth powers:

(i^4 – (i-1)^4) = 4i^3 – 6i^2 + 4i – 1

To find the sum of the cubes from 1 to n, we can sum both sides of the equation from i = 1 to i = n:
Σ(i^3) = Σ(i^4) – Σ((i-1)^4)

On the right side of the equation, we can substitute the difference between consecutive fourth powers:
Σ(i^3) = Σ(4i^3 – 6i^2 + 4i – 1)

Now we can evaluate the summation by substituting the value of the lower and upper limits:
Σ(i^3) = (4(1^3) – 6(1^2) + 4(1) – 1) + (4(2^3) – 6(2^2) + 4(2) – 1) + … + (4(n^3) – 6(n^2) + 4(n) – 1)

Simplifying the above expression, we get:
Σ(i^3) = 4(1 + 8 + 27 + … + n^3) – 6(1 + 4 + 9 + … + n^2) + 4(1 + 2 + 3 + … + n) – n

Using the formula for the sum of cubes, the first term on the right side evaluates to:
1^3 + 2^3 + … + n^3 = (n(n+1))^2/4

And, by the formula for the sum of squares, the second term on the right side evaluates to:
1^2 + 2^2 + … + n^2 = n(n+1)(2n+1)/6

And, by the formula for the sum of natural numbers, the third term on the right side evaluates to:
1 + 2 + … + n = n(n+1)/2

So, substituting these values back into our expression for Σ(i^3), we get:

Σ(i^3) = 4( (n(n+1))^2/4) – 6(n(n+1)(2n+1)/6) + 4(n(n+1)/2) – n

and combining like terms

Σ(i^3) = n^2 (n+1)^2/4

So, we have derived the formula for the sum of the cubes from 1 to n.

Solving for the sum of cubes on the right side gives us (4(1 + 8 + 27 + … + n^3))/4 – 6(1 + 4 + 9 + … + n^2) + 2(1 + 2 + 3 + … + n) – n = (n^2 * (n+1)^2) / 4