Question

if a+b+c=6 and ab+bc+ca=11 find tge value of a^3+b^3+c^3-3abc

Answer 1

Solution :

a + b + c = 6

ab + bc + ca = 11

we know,

(a + b + c)² = a² + b² + c² + 2a.b + 2bc + 2ac

6² =  a² + b² + c² + 2(ab + bc + ca)

36 = a² + b² + c² + 2(11)

a² + b² + c² = 14

a³ + b³ + c³ - 3 * a * b * c = (a + b + c)(a² + b² +c² - ab - bc - ca)

= 6(14 -  11)

= 18

a³ + b³ + c³ - 3 * a * b * c  = 18

Hope that helped

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