Combination of quadratic and cubic series I'm an eight-grader and I need help to answer this math problem (homework).

Problem:


  Calculate $$\frac{1^2+2^2+3^2+4^2+...+1000^2}{1^3+2^3+3^3+4^3+...+1000^3}$$


Attempt:

I know how to calculate the quadratic sum using formula from here: Combination of quadratic and arithmetic series but how to calculate the cubic sum? How to calculate the series without using calculator? Is there any intuitive way like previous answer? Please help me. Grazie!
We have identities for sums of powers like these. In particular:

$$1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$$

$$1^3 + 2^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4}$$

The rest is just a bit of arithmetic.
It is a quadraic sequence as 1,4,9,16,25,36 ........//

Its first term (a)=1
1st difference (d)=4-1=3
2nd difference i.e. constant difference(c)=2

There is a sum formula for any quadraic equation which is 

Sn=n/6[(n-1)3d+(n-1)(n-2)C]+an
This is for any quadraic even for sum of squares.

I have sum formula for cubic sequence.Let "b" be 1st
difference "c" be 2nd difference and "d" be 3rd diffrrence
.Also "a" is first term."n" is no. of term.Put all value in
below formula except "n" to get sum formula of any cubic
sequence.

sum formula 

Sn=n/24*[(n-1)12b+(n-1)(n-2)4c+(n-1)(n-2)(n-3)d]+an

Both formula are 100% working.