$a+b+1=0$

$a^{3} + b^{3} + c^{3} - 3abc = ( a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ac)$

$a^{3} + b^{3} + 1^{3} - 3ab.1 = (a + b + 1 )(a^{2} + b^{2} + 1 - a - b -ab)$

$a^{3} + b^{3} + 1^{3} - 3ab.1 = (0)(a^{2} + b^{2} + 1 - a - b -ab)$

$= 0$

@Vivek Vijayan Nice problem

From $a+b+1=0\quad \Rightarrow a+b = -1$

Cubing both side we have:

$(a+b)^3 = (-1)^3\quad \Rightarrow a^3 + 3a^2b + 3ab^2 + b^3 = -1$

$\Rightarrow a^3 + b^3 + 1 + 3ab(a+b) = 0\quad \Rightarrow a^3 + b^3 + 1 + 3ab(-1) = 0$

$\Rightarrow a^3 + b^3 + 1 - 3ab = \boxed{0}$

a+b=-1, so (a + b)^3 = -1, but (a+b)^3= a^3 + 3ba^2 + 3ab^2+b^3 = a^3 + b^3 + 3ab(a+b)= a^3 + b^3 -3ab= -1. Adding 1 we get 0.

$a+b=-1\\ a^2+b^2+2ab=1\\ a^2+b^2-ab=1-3ab$ Then $a^3+b^3+1-3ab\\ (a+b)(a^2+b^2-ab)+1-3ab\\ (-1)(1-3ab)+1-3ab = 0$

when a+b=(-1) then, (a+b)^2=1, then (a+b)^3=(-1 )

Acc. to formula(a+b)^2 (a*b)=1 then

Acc. To (a+b)^3=a^3+b^3+3ab(a+b)

-1=a^3+b^3+3(-1)

-1=a^3+b^3-3

a^3+b^3=2. then,

From question

a^3+b^3+1-3ab=2+1-3(1)

=3-3

=0.

Hmmm... I wonder if $a=-1$ and $b=0$ , so that the first equation is satisfied.
Now, I can substitute them in the second equation.
$a^3+b^3+1-3ab$
= $(-1)^3 + (0)^3 + 1 -3 (-1)(0)$
= $-1+0+1-0$
= $0$ . $_\square$

If a+b+c=0,then a3 +b3+c3=3abc. So,3abc-3abc=0

just take a = -1 and b= 0 .

a^3+b^3+1-3ab = a^3+b^3+3ab(a+b)+1 = (a+b)^3+1 = 0.

a+b+1=0 a+b=-1 (a+b)^3=(-1)^3 a^3+b^3+3ab(a+b)=-1 a^3+b^3+3ab(-1)=-1 [a+b=-1] a^3+b^3-3ab=-1 a^3+b^3-3ab+1=0 a^3+b^3+1-3ab=0 ans is 0

a+b+1=0 a+b=-1 (a+b)^3=(-1)^3 a^3+b^3+3ab(a+b)=-1 a^3+b^3+3ab(-1)=-1 [a+b=-1] a^3+b^3-3ab=-1 a^3+b^3-3ab+1=0 a^3+b^3+1-3ab=0 Ans is 0

-1+3ab+1-3ab=0

a+b=-1
(a+b)^3=(-1)^3
a^3+b^3+3ab(a+b)=-1
a^3+b^3+3ab(-1)=-1 [a+b=-1]
a^3+b^3-3ab=-1
a^3+b^3-3ab+1=0
a^3+b^3+1-3ab=0
Ans is 0

Given that,

a+b+1=0 a+b=-1

(a+b)^3=(-1)^3

a^3+b^3+3ab(a+b)=-1

a^3+b^3+3ab(-1)=-1 [a+b=-1]

a^3+b^3-3ab=-1

a^3+b^3-3ab+1=0

a^3+b^3+1-3ab=0

Ans is 0

a+b+1=0 a+b=-1 (a+b)^3=(-1)^3 a^3+b^3+3ab(a+b)=-1 a^3+b^3+3ab(-1)=-1 a^3+b^3-3ab=-1 a^3+b^3-3ab+1=0 a^3+b^3+1-3ab=0

a+b+c = 0 => a^3+b^3+c^3-3abc = 0 Substituting c = 1, we get the answer!

a+b+1=0 so here assume a= -2 and b=1 so above equ is satisfied now a^3= -8 and b^3=1 so -8+1+1-(3(-2)(1)) SO -6 -(-6)=0 :P

If a+b+1=0 then a+b=-1 On cubing both sides , we get, a^3 + 3ab(a+b) +b^3=-1 but a+b=-1 thus, a^3 + 3ab(-1) + b^3=-1 a^3 + b^3 +1 -3ab =0 simple and easy

replace b with -a-1 and simplify the equation, it all cancels out and gives 0 in the end

How about just plain old "a"=2 and "b"=-3 a^3 + b^3 + 1 - 3ab (2)^3 + (-3)^3 + 1 - 3ab 8 - 27 + 1 +18 -19 + 19 0

(a+b)^3=a^3+b^3+3ab(a+b) (1) now a+b=-1 and (a+b)^3=-1 hence eq(1)=a^3+b^3-3ab so eq becomes (a+b)^3+1=0

There is an identity

If a+b+c=0,

(a)^3+(b)^3+(c)^3=3abc

So,

(a)^3+(b)^3+(c)^3-3abc=0

Here,

a^3+b^3+1^3=3abc

a^3+b^3+1-3abc=0

a+b=-1, so (a + b)^3 = -1, but (a+b)^3= a^3 + 3ba^2 + 3ab^2+b^3 = a^3 + b^3 + 3ab(a+b)= a^3 + b^3 -3ab= -1. Adding 1 we get 0.

a^3 + b^3 = (a+b)^3 - 3ab(a+b) Since a+b=-1, (a+b)^3 - 3ab(a+b) = -1+3ab Since we are looking for the value of (a^3 + b^3 +1-3ab), we will substitute a^3 + b^3= -1+3ab => -1+3ab+1-3ab= 0

it is a simple indentity if sum of three numbers is zero then sum of thier cubes is three times the product of the numbers