If a = 2025, b = 2026 and c = 2027, then a³ + b³ + c³ − 3abc equals:

Question

Accepted Answer

18234. Use a³ + b³ + c³ − 3abc = ½ (a + b + c)[(a − b)² + (b − c)² + (c − a)²]. Here a + b + c = 6078 and the bracket is 1 + 1 + 4 = 6, so the value is ½ × 6078 × 6 = 6078 × 3 = 18234 — no large cubes needed.

If $a = 2025, b = 2026$ and $c = 2027$ , then $a^{3} + b^{3} + c^{3} - 3 ab c$ equals:

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If a=2025,b=2026 and c=2027, then a3+b3+c3−3abc equals:

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If $a = 2025, b = 2026$ and $c = 2027$ , then $a^{3} + b^{3} + c^{3} - 3 ab c$ equals: