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Question

If a + b + c = 9 and ab + bc + ca = 23, then a3+b3+c33abc=


A

108

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B

207

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C

669

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D

729

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Solution

The correct option is A

108


a3+b3+c33abc=(a+b+c)[a2+b2+c2(ab+bc+ca)]Now, a+b+c=9Squaring, a2+b2+c2+2(ab+bc+ca)=81 a2+b2+c2+2×23=81 a2+b2+c2+46=81 a2+b2+c2=8146=35Now ,a3+b3+c33abc=(a+b+c)[(a2+b2+c2)(ab+bc+ca)]=9[3523]=9×12=108


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