. If a + b + c = 6 and ab + bc + ca = 11, find the value of a3 +b3 +c3 − 3abc.
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7
Answer:
Given a + b + c = 6 and ab + bc + ca = 11
(a + b + c)2 = a2 + b2 +c2 +2(ab + bc + ca)
(6)2 = a2 + b2 +c2 + 2 x 11
a2 + b2 +c2 = 36 – 22 = 14
a3 +b3 +c3 − 3abc = ( a + b + c)[ a2 + b2 +c2 −(ab + bc + ca)]
= 6 x (14 - 11)
= 6 × 3
= 18
Answered by
27
Given a + b + c = 6 and ab + bc + ca = 11
(a + b + c)2 = a2 + b2 +c2 +2(ab + bc + ca)
(6)2 = a2 + b2 +c2 + 2 x 11
a2 + b2 +c2 = 36 – 22 = 14
a3 +b3 +c3 − 3abc = ( a + b + c)
[ a2 + b2 +c2 −(ab + bc + ca)]
= 6 x (14 - 11)
= 6 × 3
= 18
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