Question

ifa+b+c=0 then proove a3+b3 +c3 =3abc​

Answer 1

Step-by-step explanation:

a+b+c=0

a3+b3+c3=3abc

3(a+b+c)=3(a×b×c)

3×0=3×0

0=0

HOPE IT WILL HELP YOU

Answer 2

Answer:

Step-by-step explanation:

a+b+c=0, so (b+c)=-a,(a+c)=-b & (a+b)=-c

Now(a+b+c)^3=[(a+b)+c]^3={(a+b)^3+3(a+b)*c[(a+b)+c]+c^3=[a^3+3ab(a+b)+b^3]+3(a+b)^2*c+3(a+b)*c^2+c^3. Since a+b+c=0 we can rewrite the value of (a+b+c)^3 as under:

or,a^3+3a^2*b+3ab^2+b^3+3c(a^2+2ab+b^2)+3(a+b)*c^2+c^3=0

or,a^3+b^3+c^3+3ab(a+b)+3a^2*c+6abc+3b^2*c+3ac^2+3bc^2=0

or,a^3+b^3+c^3+3ab(a+b)+6abc+3ac(a+c)+3bc(b+c)=0, let us substitute the values of (a+b),(b+c) & (c+a) in the above equation,

a^3+b^3+c^3+3ab(-c)+6abc+3ac(-b)+3bc(-a)=0

or, a^3+b^3+c^3–3abc+6abc-3abc-3abc=0

or,a^3+b^3+c^3+6abc-9abc=0

or, a^3+b^3+c^3–3abc=0

or, a^3+b^3+c^3=3abc