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