If a+b+c=9 and a^2+b^2+c^2=35 find the value of a^3+b^3+c^3-3abc
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Given (a + b + c) = 9
Squaring on both the sides, we get
(a + b + c)2 = 92
⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 81
⇒ 35 + 2(ab + bc + ca) = 81
⇒ 2(ab + bc + ca) = 81 – 35 = 46
⇒ ab + bc + ca = 23 → (1)
Recall that a3+b3+c3-3abc = (a + b + c)( a2 + b2 + c2 – ab – bc – ca)
= 9(35 – 23)
= 9(12) = 108
Squaring on both the sides, we get
(a + b + c)2 = 92
⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 81
⇒ 35 + 2(ab + bc + ca) = 81
⇒ 2(ab + bc + ca) = 81 – 35 = 46
⇒ ab + bc + ca = 23 → (1)
Recall that a3+b3+c3-3abc = (a + b + c)( a2 + b2 + c2 – ab – bc – ca)
= 9(35 – 23)
= 9(12) = 108
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