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Question
If a + b + c = 9 and ab + bc + ca =23, then a3 + b3 + c3 − 3abc =
Options
108
207
669
729
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Solution
We have to find the value of `a^3 +b^3 +c^3 - 3abc`
Given `a+b+c = 9,ab +bc +ca = 23`
Using identity `(a+b+c)^2 = a^2 +b^2 +c^2 +2ab +2bc + 2ca` we get,
`(9)^2 = a^2 +b^2 +c^2 +2 (ab+bc +ca)`
` 9 xx 9 = a^2 +b^2 +c^2 +2 xx 23`
`81 = a^2 +b^2 +c^2 +46`
By transposing +46 to left hand side we get,
`81-46 = a^2 +b^2 +c^2`
`35 = a^2 +b^2 +c^2`
Using identity `a^3 +b^3 +c^3 -3abc = (a+b+c)[a^2 + b^2 +c^2 - (ab+bc+ca)]`
`9 xx [35 -23]`
` = 9 xx 12`
` = 108`
The value of `a^3 +b^3 +c^3 -3abc` is 108.
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