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# if a+b+c = 9 and ab+bc+ca = 23 then a cube + b cube + c cube- 3abc=

Asked by RUDRA YADAV 15th September 2013, 8:33 PM

a3+b3+c3- 3abc = (a+b+c) (a2+b2+c2-ab-bc-ca)

a+b+c= 9 (given)

ab+ bc + ca = 23 (given)

So, a3+b3+c3- 3abc = 9 (a2+b2+c2- 23)            ... (1)

(a+b+c)2= a2+b2+c2+ 2ab+ 2bc+ 2ca

92= a2+b2+c2+ 2(ab+bc+ca)

81= a2+b2+c2+ 2(23)

81 = a2+b2+c2+ 46

a2+b2+c2= 35

From (1), we get,

a3+b3+c3- 3abc = 9 (35-23) = 108

Answered by Expert 15th September 2013, 10:59 PM
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