a's and b's

at first,

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

or, $2=1-2ab$

or, $ab=$ $-\frac{1}{2}$

now,

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

or, $a^3+b^3=1+\frac{3}{2}$

or, $a^3+b^3=1+1.5=2.5$

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

$(a+b)^2=a^2+2ab+b^2$ , substituting we get

$1=2+2ab$ $\implies$ $-1=2ab$ $\implies$ $ab=-\dfrac{1}{2}$

now we have,

$a^3+b^3=1\left(2+\dfrac{1}{2}\right)=$ $\color{#D61F06}\boxed{2.5}$

From difference of second equation and share of the first 2ab=-1. So 3ab(a+b)=-3/2.