if a3+ b3 + c3 = 3abcfind the value of a + b + c
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Answered by
3
(a+b+c)³ = (a+b)³ + c ³ + 3 (a+b)² c + 3 (a+b) c²
= a³ + b³ + c³ + 3a²b + 3ab² + 3a²c + 3b²c + 6 abc + 3 ac² + 3 b c²
= 3 a b c + 3 a² b + 3 ab² + 3 a²c + 3ac² + 3 abc + 3 abc + 3 bc² + 3 cb²
= 3 a b (a+b+c) + 3 ac (a+b+c) + 3 bc ( a+b+c)
= (a+b+c) (ab+bc+ca) 3
(a+b+c)² = 3 (ab+bc+ca)
===========================
a+b+c = k
k² = 3 a b + 3 (k-c) c
k² - 3 k c + 3c² - 3 ab = 0
k = [ 3c +- √{12 ab - 3c² } ] / 2
= a³ + b³ + c³ + 3a²b + 3ab² + 3a²c + 3b²c + 6 abc + 3 ac² + 3 b c²
= 3 a b c + 3 a² b + 3 ab² + 3 a²c + 3ac² + 3 abc + 3 abc + 3 bc² + 3 cb²
= 3 a b (a+b+c) + 3 ac (a+b+c) + 3 bc ( a+b+c)
= (a+b+c) (ab+bc+ca) 3
(a+b+c)² = 3 (ab+bc+ca)
===========================
a+b+c = k
k² = 3 a b + 3 (k-c) c
k² - 3 k c + 3c² - 3 ab = 0
k = [ 3c +- √{12 ab - 3c² } ] / 2
Answered by
4
If a3+b3+c3 =3abc then a3+b3+c3-3abc=0
=> (a+b+c)(a2+b2+c2-ab-bc-ca)=0 =>
(a+b+c)=0 or ( a2+b2+c2 -ab-bc-ca)=0
=> (a+b+c)=0
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