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(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

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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