Let's create an identity

First let us get acquainted to this identity.

${ a }^{ 4 }+{ b }^{ 4 }+{ c }^{ 4 }+{ d }^{ 4 }+({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 })(ab+ac+ad+bc+bd+cd)+4abcd=(a+b+c+d)({ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }+{ d }^{ 3 }+abc+abd+bcd+acd)$

Now just substitute the values of w,x,y and z to get the answer.

Now, how is this identity derived? Don't worry it's derivation is not as lengthy as the identity. Let p(x) be a biquadratic polynomial with a roots a,b,c and d

$\therefore \quad p(x)={ x }^{ 4 }-(a+b+c+d){ x }^{ 3 }+(ab+ac+ad+bc+bd+cd){ x }^{ 2 }-(abc+abd+bcd+acd)x+abcd$

$\therefore \quad p(a)={ a }^{ 4 }-(a+b+c+d){ a }^{ 3 }+(ab+ac+ad+bc+bd+cd)a^{ 2 }-(abc+abd+bcd+acd)a+abcd=0$

$\therefore \quad p(b)={ b }^{ 4 }-(a+b+c+d)b^{ 3 }+(ab+ac+ad+bc+bd+cd)b^{ 2 }-(abc+abd+bcd+acd)b+abcd=0$

$\therefore \quad p(c)={ c }^{ 4 }-(a+b+c+d)c^{ 3 }+(ab+ac+ad+bc+bd+cd)c^{ 2 }-(abc+abd+bcd+acd)c+abcd=0$

$\therefore \quad p(d)={ d }^{ 4 }-(a+b+c+d)d^{ 3 }+(ab+ac+ad+bc+bd+cd)d^{ 2 }-(abc+abd+bcd+acd)d+abcd=0$

adding all the results yields the above identity.

Using Newton's sum, we have:

$\begin{aligned} a^4+b^4+c^4 +d^4 & = (a+b+c+d)(a^3+b^3+c^3 +d^3) \\ & \quad - (ab+ac+ad+bc+bd+cd )(a^2+b^2+c^2 +d^2) \\ & \quad + (abc+abd+acd+bcd)(a+b+c +d) - 4abcd \\ \Rightarrow w & = (a+b+c+d)(a^3+b^3+c^3 +d^3) - yx \\ & \quad + (abc+abd+acd+bcd)(a+b+c +d) - 4z \end{aligned}$

$\Rightarrow (a+b+c+d)(a^3+b^3+c^3 +d^3 + abc+abd+acd+bcd) \\ \quad \quad =\boxed{w +xy + 4z}$

Let $a^n+b^n+c^n+d^n=P_n$ , $a+b+c+d=S_1$ , $ab+ac+ad+bc+bd+cd=S_2$ , $abc+abd+acd+bcd=S_3$ and $abcd=S_4$ . By Newton's Sums we know that:

$P_1=S_1 \\ P_2=S_1^2-2S_2 \\ P_3=S_1^3-3S_1S_2+3S_3 \\ P_4=S_1^4-4S_1^2S_2+4S_1S_3+2S_2^2-4S_4$

Then let $A$ the expression that we want, that is $A=S_1(P_3+S_3)$ . Substituting the values obtained we obtain: $A=S_1(S_1^3-3S_1S_2+3S_3+S_3)=S_1^4-3S_1^2S_2+4S_1S_3$ .

Now, substituting the given values we obtain:

$x=S_1^2-2y \\ w=S_1^4-4S_1^2y+4S_1S_3+2y^2-4z \\ A=S_1^4-3S_1^2y+4S_1S_3$

Now, subtract the third equation with the second one:

$A-w=S_1^2y-2y^2+4z$

Factor some terms and substitute the first equation:

$A-w=y(S_1^2-2y)+4z \\ A-w=y(x)+4z$

Finally, solve for $A$ :

$A=\boxed{w+xy+4z}$