$x^7+y^7+x^4y^3+x^3y^4$

$x^7+y^7+x^4y^3+x^3y^4 = \left(x^3+y^3\right)\left(x^4+y^4\right)$ $x^7+y^7+x^4y^3+x^3y^4 = \left(x+y\right)\left(x^2+y^2-xy\right)\left(\left(x^2+y^2\right)^2-2x^2y^2\right)$

From the given values,
$x+y =1$ $x^2+y^2 = 2$ $\Rightarrow xy = \frac{-1}{2}$

Using all these values,
$x^7+y^7+x^4y^3+x^3y^4 = \frac{5}{2}\cdot\frac{7}{2} = \boxed{\frac{35}{4}}$

First, note that $(x+y)^2=x^2+2xy+y^2=1\implies xy=-\dfrac{1}{2}$

Now let us define $x,y$ as the roots of a quadratic. This quadratic, by Vieta's, is $x^2-x-\dfrac{1}{2}$ .

Now, notice that $x^7+y^7+x^4y^3+x^3y^4=(x^3+y^3)(x^4+y^4)$ . We want to find the values of $x^3+y^3$ and $x^4+y^4$ .

Now define $P_n=x^n+y^n$ .

By Newton's Sums, $P_3-P_2-\dfrac{1}{2}P_1=0$ ; we substitute $P_1=1$ and $P_2=2$ (from the conditions $x+y=1$ and $x^2+y^2=2$ ) to get $P_3=\dfrac{5}{2}$ .

By Newton's Sums, $P_4-P_3-\dfrac{1}{2}P_2=0$ ; we substitute $P_2=2$ and $P_3=\dfrac{5}{2}$ to get $P_4=\dfrac{7}{2}$ .

We want to find $P_3P_4$ , which is simply $\dfrac{5}{2}\cdot\dfrac{7}{2}=\boxed{\dfrac{35}{4}}$ and we are done. $\Box$ .

Sidenote: using Newton's Sums was a little bit overkill on this problem, but if they asked you to compute $x^{12}+y^{12}+x^5y^7+x^7y^5$ , then this strategy would be much easier.

We know that $x+y=1$ and $x^2+y^2=2$

$x^7+y^7+x^4y^3+x^3y^4=\\ =x^4(x^3+y^3)+y^4(x^3+y^3)=\\ =(x^4+y^4)(x^3+y^3)$

Note that $x+y=1\\ (x+y)^2=1\\ x^2+y^2+2xy=1\\ 2xy+2=1\\ xy= - \dfrac{1}{2}$

Now we calculate each of the factors:

$(x^2+y^2)(x+y)=2\cdot 1\\ x^3+y^3+x^2y+xy^2=2\\ x^3+y^3+xy(x+y)=2\\ x^3+y^3+(-\dfrac{1}{2}\cdot 1)=2\\ x^3+y^3-\dfrac{1}{2}=2\\ x^3+y^3=\dfrac{5}{2}$

$(x^3+y^3)(x+y)=\dfrac{5}{2}\cdot 1\\ x^4+y^4+x^3y+xy^3=\dfrac{5}{2}\\ x^4+y^4+xy(x^2+y^2)=\dfrac{5}{2}\\ x^4+y^4+(-\dfrac{1}{2}\cdot 2)=\dfrac{5}{2}\\ x^4+y^4-1=\dfrac{5}{2}\\ x^4+y^4=\dfrac{5}{2}+1\\ x^4+y^4=\dfrac{7}{2}$

So

$x^7+y^7+x^4y^3+x^3y^4=(x^4+y^4)(x^3+y^3)=\dfrac{7}{2}\cdot \dfrac{5}{2}=\dfrac{35}{4}$