Can YOU Solve This Cubic Equation?

[@Quest3669]

Solving gives ..t^3+20t-225=0 or t= 5 for real soln. where t= x+6= 5 or x = -1 soln.

[@StaR-uw3dc]

Happy New Year 2025 = (20+25)(20+25)

[@SKASKumar]

Happy new year

[@neerajgupta7856]

-1

[@gaiatetuya92]

ｘ＝ー１のとき（左辺) ＝１^3＋2^3＋3^3＋・・・＋9^3＝（9×10/2)^2 = 45^2 = 2025 ＝(右辺） よって x=ー１

[@Fjfurufjdfjd]

χ=-1

[@潘博宇-k4l]

X=-1.

[@michaeldoerr5810]

I actually have learned of something: Given that one of the properties of 2025 is that it is equal to 1^3+2^3+3^3+...+9^3, compare that to this series, this indicates that x = -1. I think that it does. I could be wrong.

[@Fjfurufjdfjd]

Θα το κρατησω αυτο.

[@michaeldoerr5810]

@Fjfurufjdfjd Could you please translate??? I hope that that was a compliment!!!

[@Fjfurufjdfjd]

Ειναι οντως μια διαπιστωση. Εγω στην αποδειξη μου χρησιμοποιησα τους τυπους 1^2+2^2+...+n^2=[n(2n+1)(n+1)]/6 και 1^3+2^3+...+n^3=[n(n+1)/2]^2 που αποδεικνυονται ευκολα με την επαγωγικη μεθοδο.

[@michaeldoerr5810]

@@Fjfurufjdfjd I also think that this is a number theoretic technique that is more precise than the inductive method. I could be wrong and I hope that I am on to something.

[@morriswangmonroewine2128]

Happy 2025 , Every One !!! ❤❤❤ 🎉🎉🎉

[@SidneiMV]

1³ + 2³ + .. + (x + 10)³ = (x + 10)²(x + 11)²/4 1³ + 2³ + .. + (x + 1)³ = (x + 1)²(x + 2)²/4 (x + 1)²(x + 2)²/4 + 2025 = (x + 10)²(x + 11)²/4 (x² + 21x + 110)² - (x² + 3x + 2)² = 8100 (2x² + 24x + 112)(18x + 108) = 8100 36(x² + 12x + 56)(x + 6) = 8100 (x² + 12x + 56)(x + 6) = 225 x + 6 = u => x = u - 6 u² = x² + 12x + 36 (u² + 20)u = 225 u³ + 20u - 225 = 0 u³ - 125 + 20(u - 5) = 0 (u - 5)(u² + 5u + 45) = 0 u = 5 => x + 6 = 5 => *x = -1*

[@karthikmalasani]

If x was a real number but not an integer, then your first two sums would not have a difference of 2025.