This is a rather encumbered solution for a not-as-hard problem, and it is the only way I see to make sense of what your friend is saying. On the other hand, it is cool that it let us know about Newton's identities and their proof using generating functions.
Let us form the formal series \begin{align}F(x,y,z,t)&:=(x+y+z)t+(z^2+y^2+z^2)t^2+(x^3+y^3+z^3)t^3+\ldots\\&=3t+5t^2+7t^3+at^4+\ldots,\end{align}
where $a:=x^4+y^4+z^4$ is the number we want to find.
Forming this power series is kind of adding the given equations (as your friend suggested), but we multiply them first by $t,t^2,t^3,...$, respectively, before adding.
Consider the polynomial \begin{align}G(x,y,z,t)&:=-1+(x+y+z)t+(xy+xz+yz)t^2-(xyz)t^3\\&=-1+3t+(xy+xz+yz)t^2-(xyz)t^3\end{align}
We multiply $F$ times $G$. It happens that we get a polynomial:
\begin{align}
FG&=\left(\sum_{k=1}^{\infty}(xt)^k+\sum_{k=1}^{\infty}(yt)^k+\sum_{k=1}^{\infty}(zt)^k\right)G\\
&=-\left(\frac{xt}{1-xt}+\frac{yt}{1-yt}+\frac{yt}{1-xt}\right)(1-xt)(1-yt)(1-zt)\\
&=t\left((-x)(1-yt)(1-zt)+(-y)(1-xt)(1-zt)+(-z)(1-xt)(1-yt)\right)\\
&=-(x+y+z)t+(xy+xz+yz)t^2-(xyz)t^3\\
&=-3t+(xy+xz+yz)t^2-(xyz)t^3
\end{align}
In the second equality we used the sum of the geometric series more or less like your friend said.
Comparing coefficients of the multiplication $FG$ with the result in the last line we get (the Newton's identities for the values in this problem)
\begin{align}
2(xy+xz+yz)&=9-5\\
3xyz&=3(xy+xz+yz)-15+7\\
0&=3(xyz)-5(xy+xz+yz)+21-(x^4+y^4+z^4)
\end{align}
From these, substituting from top to bottom, we find the value of $x^4+y^4+z^4$. Notice that the method allows us to continue using the equalities between coefficients of higher degree in $t$ to compute $x^5+y^5+z^5$, $x^6+y^6+z^6$, and so on ...
Use
$$\cos36 = \frac {\sin108}{2\sin36}
= \frac {\sin36+2\sin36\cos72}{2\sin36}=\frac12+\cos72$$
to factorize the equation as follows
$$\begin{align}
& \sin 84\sin(54-x)-\sin 54 \sin x \\
& =\cos 6 \cos 36 \cos x - ( \cos 36+\cos 6 \sin 36 )\sin x \\
& =\frac12(\cos30+\cos 42 ) \cos x
- \left(\frac12 + \cos 72 + \frac12(\sin 42+ \sin30 )\right)\sin x \\
& =\frac12\left(\frac{\sqrt3}2+\cos 42 \right) \cos x
- \left(\frac34 +\cos (42+30) + \sin30\sin 42\right)\sin x \\
& =\frac12\left(\frac{\sqrt3}2+\cos 42 \right) \cos x
- \left(\frac34 + \frac{\sqrt3}2\cos42 \right)\sin x \\
& =\frac12\left(\frac{\sqrt3}2+\cos 42 \right) (\cos x -\sqrt3 \sin x) = 0
\end{align}$$
Thus,
$$\tan x = \frac1{\sqrt3}$$
and the angle in the source problem is $30^\circ$.
Best Answer
Consider Heron's formula: the area of a triangle with sides $a, b, \text{and } c$ is
$$ \sqrt{s(s-a)(s-b)(s-c)} $$
where $s$ is the semi-perimeter $\frac12 (a + b + c)$.
Let $a, b, \text{and } c$ be $\sqrt{5}, \sqrt{6}, \text{and } \sqrt{7}$. Then the area is the square root of your expression divided by $4$. So, what is the area of this triangle? Use the law of cosines to find the cosine of the angle $C$ opposite $c$:
$$ \begin{align} 7 &= 5 + 6 - 2 \sqrt{5}\sqrt{6}\cos{C}\\ 2\sqrt{30}\cos{C} &= 4\\ \cos{C} &= \frac{2}{\sqrt{30}} \end{align} $$
But the area of the triangle is $\frac12 ab\sin{C}$.
$$ \frac12 ab\sin{C} = \frac12 \sqrt{30} \frac{\sqrt{26}}{\sqrt{30}} = \frac12\sqrt{26}. $$
Your expression is therefore the square of $2\sqrt{26}$, which is $104$.