Find the value of $x_1^6 +x_2^6$ of this quadratic equation without solving it I got this question for homework and I've never seen anything similar to it.
Solve for $x_1^6+x_2^6$ for the following quadratic equation where $x_1$ and $x_2$ are the two real roots and $x_1 > x_2$, without solving the equation.
$25x^2-5\sqrt{76}x+15=0$
I tried factoring it and I got $(-5x+\sqrt{19})^2-4=0$
What can I do afterwards that does not constitute as solving the equation? Thanks.

\begin{align}
x_1^6+x_2^6 &= (x_1^2+x_2^2)^3-3x_1^4x_2^2-3x_1^2x_2^4 \\
&= (x_1^2+x_2^2)^3-3(x_1x_2)^2(x_1^2+x_2^2).
\end{align}
Since $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2$, therefore:
\begin{align}
x_1^6+x_2^6 &= ((x_1+x_2)^2-2x_1x_2)^3-3(x_1x_2)^2((x_1+x_2)^2-2x_1x_2) \\
&= \left( \left(\frac{5\sqrt{76}}{25}\right)^2 -2\left(\frac{15}{25}\right) \right)^3 -3\left(\frac{15}{25}\right)^2 \left( \left(\frac{5\sqrt{76}}{25}\right)^2 -2\left(\frac{15}{25}\right) \right).
\end{align}

The values of $x_1x_2$ and $x_1+x_2$ come from the following argument:
\begin{align}
25(x-x_1)(x-x_2) &= 25x^2-25(x_1+x_2)x+25x_1x_2 \\
&= 25x^2-5\sqrt{76}+15.
\end{align}
Now equate the cofficents of both polynomials to get the values of $x_1x_2$ and $x_1+x_2$ .
