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Question
If a + b + c = 9 and a2+ b2 + c2 =35, find the value of a3 + b3 + c3 −3abc
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Solution
In the given problem, we have to find value of a3 + b3 + c3 −3abc
Given a + b + c = 9 , a2+ b2 + c2 =35
We shall use the identity
`(a+b+c)^2 = a^2 +b^2 + c^2 + 2 (ab+bc+ ca)`
`(a+b+c)^2 =35 + 2 (ab+bc+ ca)`
`(9)^2 =35 + 2 (ab+bc+ ca)`
`81 - 35 = 2 (ab+bc+ ca)`
`46/ 2 = (ab+bc+ ca)`
`23 = (ab+bc+ ca)`
We know that
`a^3 + b^3 + c^3- 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc -ca)`
`a^3 + b^3 + c^3- 3abc = (a+b+c)[(a^2 + b^2 + c^2) - (ab + bc +ca)`
Here substituting `a+b+c = 9,a^2 +b^2 + c^2 = 35 , ab +bc + ca = 23` we get
`a^3 +b^3 + c^3 - 3abc = 9 [(35 - 23)]`
` =9 xx 12`
` = 108`
Hence the value of a3 + b3 + c3 −3abc is 108.
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