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Question

If a + b + c = 9 and ab + bc + ca = 26, find the value of a3+b3+c33abc.


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Solution

a + b + c = 9, ab + bc + ca = 26

Squaring, we get

(a+b+c)2=(9)2a2+b2+c2+2(ab+bc+ca)=81 a2+b2+c2+2×26=81 a2+b2+c2+52=81 a2+b2+c2=8152=29Now, a3+b3+c33abc=(a+b+c)[(a2+b2+c2(ab+bc+ca))]=9[2926]=9×3=27


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