1.

If a+b+c=15, a²+b²+c²=83. Then find a³+b³+c³-3abc

Answer»

A^2 + b^2 + c^2 = 83 ------(1)

a + b+ c = 15

On squaring both sides, we get

a^2 + b^2 + c^2 + 2( ab + bc + CA) = 225

=> 83 + 2 ( ab + bc + ca) = 225

=> 2 ( ab + bc + ca) = 142

=> ab + bc + ca = 71 --------(2)



Now,

a^3 + b^3 + c^3 - 3abc = ( a +b + c) (a^2 + b^2 + c^2 - ab - bc - ca)

= ( 15) [ 83 - (ab + bc + ca)]

= (15)(83-71)

= 15 × 12

= 180


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