Square, square , and again

First, find $a$ from the expression

$a + b = 1$

$a = 1 - b$

Now lets find $b$

$a^2 + b^2 = 5$

$= (1-b)^2 + b^2 = 5$

$= 1 - 2b + b^2 + b^2= 5$

$=2b^2 - 2b - 4 = 0$

$= b^2 - b - 2 = 0$

$(b-2)(b+1)$ = 0

$b = 2$ or $b=-1$

Now to the result

$b=2$ - - - - - - - - - - - - - $a = 1-2 = -1$ - - - - - - - - - - - - - - - $( -1)^3 + 2^3 = 7$

$b=-1$ - - - - - - - - - - - - - $a = 1+1 = 2$ - - - - - - - - - - - - - - - $2^3 + (-1)^3 = 7$

The answer is 7

Usually before solving I prefer to summon my inner Sherlock Holmes (yeah I'm overreacting :D) and do some investigation.

$a + b = 1$

$a^2 + b^2 = 5$

There actually exists a relation between these two equations!

$(a + b)^2 = a^2 + 2ab + b^2$

By substituting with the given values:

$1 = 5 + 2ab$

$ab = -2$ (Important for later use)

And we already know that $a^3 + b^3 = (a + b) (a^2 - ab +b^2)$

I think everything looks easy now;

$a^3 + b^3 = (1) (5 - (-2))$

$a^3 + b^3 = \boxed{7}$

$a + b =1 \Rightarrow (a + b)^2 = 1 = a^2 + b^2 + 2ab = 5 + 2ab \Rightarrow$ $ab = - 2 \Rightarrow (a + b)^3 = 1 = a^3 + b^3 + 3ab(a + b) \Rightarrow a^3 + b^3 - 6 = 1 \Rightarrow a^3 + b^3 = 7.$ This is the on ly possibility for $a^3 + b^3$ so sum of all possiblities is $\boxed{7}$

(a^2 + b^2)( a + b ) = 1 x 5 = 5 expanding that's a^3 + a^2b + ab^2 + b^3 = 5 a^3 + b^3 = 5 - a^2b + ab^2 simplify a^3 + b^3 = 5 - ab(a + b ) -----------------------------------eq. 1 but ( a + b )^2 = 1 ab = -2 and ( a + b ) =1 substitute to eq. 1 a^3 + b^3 = 5 - -(2) = 7 answer. cheeeeeeeeeeerrrrrrrrrrrrrrrrsssssssssssss