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If a + b + c = 6 and ab + bc + ca = 11, find the value of a^3 +b^3 +c^3 − 3abc.

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Given a + b + c = 6 and ab + bc + ca = 11

(a + b + c)2 = a+ b2 +c2 +2(ab + bc + ca) 
(6)= a+ b2 +c2 + 2 x 11
a+ b2 +c2 = 36 – 22 = 14 
a+b+c− 3abc = ( a + b + c)[ a+ b2 +c2 −(ab + bc + ca)] 
                              = 6 x (14 - 11)

                              = 6 × 3

                              = 18

Posted by

Ravindra Pindel

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