If a + b + c = 6, a² + b² + c² = 14 and abc = 6, then a³ + b³ + c³ equals:

Question

Accepted Answer

36. First ab + bc + ca = [(a + b + c)² − (a² + b² + c²)]/2 = (36 − 14)/2 = 11. Then a³ + b³ + c³ = 3abc + (a + b + c)(a² + b² + c² − ab − bc − ca) = 18 + 6(14 − 11) = 18 + 18 = 36. (The numbers are 1, 2, 3, and 1 + 8 + 27 = 36.)

If $a + b + c = 6, a^{2} + b^{2} + c^{2} = 14$ and $ab c = 6$ , then $a^{3} + b^{3} + c^{3}$ equals:

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If a+b+c=6,a2+b2+c2=14 and abc=6, then a3+b3+c3 equals:

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If $a + b + c = 6, a^{2} + b^{2} + c^{2} = 14$ and $ab c = 6$ , then $a^{3} + b^{3} + c^{3}$ equals: