+130466 Let n be the first number and (n + 1) be the second...so we have
n^3 + (n + 1)^3 = 341 expand
n^3 + n^3 + 3n^2 + 3n + 1 = 341 simplify
2n^3 + 3n^2 + 3n - 340 = 0
Using the Factor Theorem here might be a little difficult.....using the on-site solver is easier.....!!
2[n3]+3[n2]+3n−340=0⇒{n=−(5×√15×i+13)4n=(5×√15×i−13)4n=5}⇒{n=−(134+4.8412291827602082i)n=−134+4.8412291827602082in=5}
So n = 5 is the only "real" answer ...and n+1 = 5+1 = 6
Check
5^3 + 6^3 =
125 + 216 =
341 ......√√√
