We have:

\(a^3+b^3=\left(a+b\right)\left(a^2+ab+b^2\right)=2\) (*)

Other way:

\(a^2+ab+b^2=a^2+2.\dfrac{1}{2}b+\left(\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\)

\(=\left(a+\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\)

Because \(\left(a+\dfrac{1}{2}b\right)^2\ge0\) with \(\forall a,b\)

\(\dfrac{3}{4}b^2\ge0\) with \(\forall b\)

So \(\left(a+\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\ge0\) with \(\forall a,b\)

or \(a^2+ab+b^2\ge0\) with \(\forall a,b\) (**)

From (*) and (**) we have  \(a+b\le2\)

Your ex is so hard to do :) 

We have:

a3+b3=(a+b)(a2+ab+b2)=2a3+b3=(a+b)(a2+ab+b2)=2 (*)

Other way:

a2+ab+b2=a2+2.12b+(12b)2+34b2a2+ab+b2=a2+2.12b+(12b)2+34b2

=(a+12b)2+34b2=(a+12b)2+34b2

Because (a+12b)2≥0(a+12b)2≥0 with ∀a,b∀a,b

34b2≥034b2≥0 with ∀b∀b

So (a+12b)2+34b2≥0(a+12b)2+34b2≥0 with ∀a,b∀a,b

or a2+ab+b2≥0a2+ab+b2≥0 with ∀a,b∀a,b (**)

From (*) and (**) we have  a+b≤2a+b≤2

Your ex is so hard to do :)