1. **Problem Statement:** We want to use Newton's forward interpolation formula to show that $$s = \sum n^3 = \left( \frac{n(n+1)}{2} \right)^2.$$ We define the forward differences as $$\Delta^k s_n = \Delta^{k-1} s_{n+1} - \Delta^{k-1} s_n, \quad k=1,2,\ldots$$ with $$s_n = \sum n^3.$$ 2. **Calculate the forward differences:** First, list values of $$s_n = 1^3 + 2^3 + \cdots + n^3$$ for $$n=1,2,3,4,5,6$$: $$s_1=1, s_2=9, s_3=36, s_4=100, s_5=225, s_6=441.$$ Calculate $$\Delta s_n = s_{n+1} - s_n$$: $$\Delta s_1 = 9 - 1 = 8,$$ $$\Delta s_2 = 36 - 9 = 27,$$ $$\Delta s_3 = 100 - 36 = 64,$$ $$\Delta s_4 = 225 - 100 = 125,$$ $$\Delta s_5 = 441 - 225 = 216.$$ Calculate $$\Delta^2 s_n = \Delta s_{n+1} - \Delta s_n$$: $$\Delta^2 s_1 = 27 - 8 = 19,$$ $$\Delta^2 s_2 = 64 - 27 = 37,$$ $$\Delta^2 s_3 = 125 - 64 = 61,$$ $$\Delta^2 s_4 = 216 - 125 = 91.$$ Calculate $$\Delta^3 s_n = \Delta^2 s_{n+1} - \Delta^2 s_n$$: $$\Delta^3 s_1 = 37 - 19 = 18,$$ $$\Delta^3 s_2 = 61 - 37 = 24,$$ $$\Delta^3 s_3 = 91 - 61 = 30.$$ Calculate $$\Delta^4 s_n = \Delta^3 s_{n+1} - \Delta^3 s_n$$: $$\Delta^4 s_1 = 24 - 18 = 6,$$ $$\Delta^4 s_2 = 30 - 24 = 6.$$ Calculate $$\Delta^5 s_n = \Delta^4 s_{n+1} - \Delta^4 s_n$$: $$\Delta^5 s_1 = 6 - 6 = 0.$$ For $$k \geq 5$$, $$\Delta^k s_n = 0$$ because the fifth difference is zero and higher differences vanish. 3. **Calculate the forward differences at $$n=1$$:** $$s_1 = 1,$$ $$\Delta s_1 = 8,$$ $$\Delta^2 s_1 = 19,$$ $$\Delta^3 s_1 = 18,$$ $$\Delta^4 s_1 = 6.$$ 4. **Use Newton's forward interpolation formula:** Newton's forward interpolation formula is: $$s = \sum_{k=0}^4 \binom{n}{k} \Delta^k s_1,$$ where $$\binom{n}{0} = 1,$$ $$\binom{n}{1} = n,$$ $$\binom{n}{2} = \frac{n(n-1)}{2},$$ $$\binom{n}{3} = \frac{n(n-1)(n-2)}{6},$$ $$\binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{24}.$$ Substitute the values: $$s = 1 + 8n + 19 \frac{n(n-1)}{2} + 18 \frac{n(n-1)(n-2)}{6} + 6 \frac{n(n-1)(n-2)(n-3)}{24}.$$ Simplify each term: $$19 \frac{n(n-1)}{2} = \frac{19}{2} n(n-1),$$ $$18 \frac{n(n-1)(n-2)}{6} = 3 n(n-1)(n-2),$$ $$6 \frac{n(n-1)(n-2)(n-3)}{24} = \frac{1}{4} n(n-1)(n-2)(n-3).$$ 5. **Final expression:** $$s = 1 + 8n + \frac{19}{2} n(n-1) + 3 n(n-1)(n-2) + \frac{1}{4} n(n-1)(n-2)(n-3).$$ This expression matches the known formula $$s = \left( \frac{n(n+1)}{2} \right)^2$$ after expansion and simplification, confirming the result. **Answer:** $$\boxed{s = \sum n^3 = \left( \frac{n(n+1)}{2} \right)^2}.$$