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Question -

If a + b + c = 9, and a2 + b2 + c2 = 35, find the value of a3 + b3 + c3 – 3abc.



Answer -

a + b +c = 9
Squaring, we get
(a + b + c)2 = (9)2
 a2 + b2 + c2 + 2 (ab + be + ca)= 81
 35 + 2(ab + bc + ca) = 81
2(ab + bc + ca) = 81 – 35 = 46
 ab + bc + ca = 462 =23
Now, a3 + b3 + c3 – 3abc
= (a + b + c) [a2 + b2 + c2 –(ab + bc + ca)]
= 9[35 – 23] = 9 x 12 = 108

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