Question

prove that a+b+c=a3+b3+c3=3abc​

Answer 1

We know,

We know,a

We know,a 3

We know,a 3 +b

We know,a 3 +b 3

We know,a 3 +b 3 +c

We know,a 3 +b 3 +c 3

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)Putting a+b+c=0 on RHS, we get,

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)Putting a+b+c=0 on RHS, we get,a

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)Putting a+b+c=0 on RHS, we get,a 3

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)Putting a+b+c=0 on RHS, we get,a 3 +b

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)Putting a+b+c=0 on RHS, we get,a 3 +b 3

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)Putting a+b+c=0 on RHS, we get,a 3 +b 3 +c

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)Putting a+b+c=0 on RHS, we get,a 3 +b 3 +c 3

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)Putting a+b+c=0 on RHS, we get,a 3 +b 3 +c 3 −3abc=0

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)Putting a+b+c=0 on RHS, we get,a 3 +b 3 +c 3 −3abc=0⟹a

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)Putting a+b+c=0 on RHS, we get,a 3 +b 3 +c 3 −3abc=0⟹a 3

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)Putting a+b+c=0 on RHS, we get,a 3 +b 3 +c 3 −3abc=0⟹a 3 +b

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)Putting a+b+c=0 on RHS, we get,a 3 +b 3 +c 3 −3abc=0⟹a 3 +b 3

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)Putting a+b+c=0 on RHS, we get,a 3 +b 3 +c 3 −3abc=0⟹a 3 +b 3 +c

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)Putting a+b+c=0 on RHS, we get,a 3 +b 3 +c 3 −3abc=0⟹a 3 +b 3 +c 3

We know,a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −ab−bc−ca)Putting a+b+c=0 on RHS, we get,a 3 +b 3 +c 3 −3abc=0⟹a 3 +b 3 +c 3 =3abc, Hence proved.