No, I'm afraid it isn't 3

a+b=1,a^2+b^2=2

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

1=2+2ab

ab=-1/2

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

by substituting the values we get a^3+b^3= 5/2

so 5+2=7

Let's start by multiplying both equations by each other. $(a+b)(a^{2}+b^{2})$ expands to $a^{3}+ab^{2}+ba^{2}+b^{3}$ . When we factor the middle chunk of the expression ( $ab^{2}+ba^{2}$ ) we get $ab(a+b)$ . Because we already know that $a+b=1$ , we see that all that we don't know is the $ab$ . To find this, expand $(a+b)^{2}$ , and get $a^{2}+2ab+b^{2}$ . We already know $a^{2}+b^{2}=2$ , and subtracting that from the whole equation, we find that $2ab=-1$ . Therefore, $ab=\frac{-1}{2}$ . Plugging this in for the original expansion of $(a+b)(a^{2}+b^{2})$ gets $a^{3}-\frac{1}{2}+b^{3}=2$ . Solving, we see that $a^{3}+b^{3}=\frac{5}{2}$ . Adding $5+2$ gets the answer $\boxed{7}$ . That is a heck lot of LaTeX.

First of all we should know the following formulas to solve this : $(x+y)^3=x^3+y^3+3xy(x+y)\quad \text{and} \quad x^2+y^2=(x+y)^2-2xy=(x-y)^2+2xy$

Given that, $a+b=1...(i)\quad a^2+b^2=2....(ii)$

From (ii), we get the following ----

$a^2+b^2=2$

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

$\implies 1^2-2ab=2$ [from (i)]

$\implies 2ab=(-1) \implies ab=\boxed{(\frac{-1}{2})}$

Now, we cube the eq. (i) and putting the value of $ab$ , we get---

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

$\implies a^3+b^3+3ab(a+b)=1$

$\implies a^3+b^3+3ab=1$ [from (i)]

$\implies a^3+b^3+3(\frac{-1}{2})=1$ [since we found that $ab=(\frac{-1}{2})$ ]

$\implies a^3+b^3-\frac{3}{2}=1$

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

$\implies a^3+b^3=\boxed{\frac{5}{2}}$

So, now $a^3+b^3$ is in the form of $\frac{x}{y}$ with $x,y$ as coprime integers i.e., $x=5,y=2$

Then, the reqd. sum $=x+y=5+2=\boxed{7}$

we can write a^2 + b^2 as (a+b)^2 -2ab and then calculate ab = -1/2. Then we can write a^3 +b^3 as (a+b)(a^2+b^2 - ab) and then on putting the values we get 5/2 so answer is 7!

first we have to find out ab ..... ab= -1/2

then we have to put it in the formula of a^3+b^3 which is equal to 5/2......therefore 5+2=7

$a^3+b^3=(a+b)(a^2-ab+b^2)=(1)(a^2-ab+b^2)=a^2+b^2-ab=\frac{3}{2}(a^2+b^2)-\frac{1}{2}(a^2+b^2)-ab=\frac{3}{2}(a^2+b^2)-\frac{1}{2}(a^2+2ab+b^2)$ $=\frac{3}{2}(a^2+b^2)-\frac{1}{2}(a+b)^2=\frac{3}{2}(2)-\frac{1}{2}(1)^2=3-\frac{1}{2}=\frac{5}{2}$

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

Since $a+b=1$ and $a^{2}+b^{2}=2$ , $a^{3}+b^{3}=2-ab.$

Now we need to find $ab$ .

Note that $(a+b)^{2}=a^{2}+b^{2}+2ab=2+2ab=1 \implies ab=-\frac{1}{2}$ , and the desired fraction is $2-(-\frac{1}{2})=\frac{5}{2}$ , which yields the answer $\boxed{7}$ .

a+b = 1, a2+b2=2 so ab = -1/2, now a3+b3 = (a+b)(a2-ab+b2) = 1(2+1/2) = 5/2 = x/y so x+y = 5+2=7

a+b=1, a^2+b^2 =2 Now (a+b)^2=1^2=1 = a^2 + b^2 +2ab=1 = 2+2ab=1 =2ab= -1 ab= -1/2 a^3 + b^3 = (a+b)3- 3ab(a+b) a^3 +b^3 = 1+3/2 a^3 + b^3 = 5/2 x+y=7

a+b=1 squaring on both sides.
(a+b)^2=1^2. a^2+b^2+2ab=1 2+2ab=1==>2ab=1- 2==>>ab= -1/2 (given a^2+b^2 =2). a^3+b^3=(a+b)(a^2+b^2-ab)
=(1)(2-(-1/2)) =2+1/2
= 5/2
answer=5+2=7