Question

Solve the equations:
$x+y+z+u=0$,
$ax+by+cz+du=0$,
$a^{2}x+b^{2}y+c^{2}z+d^{2}u=0$,
$a^{3}x+b^{3}y+c^{3}z+d^{3}u=k$.

Answer 1

$\Delta =\left|\begin{array}{cccc}1& 1& 1& 1\\ a& b& c& d\\ a^2& b^2& c^2& d^2\\ a^3& b^3& c^3& d^3\end{array}\right|=(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$

${\Delta }_1=\left|\begin{array}{cccc}0& 1& 1& 1\\ 0& b& c& d\\ 0& b^2& c^2& d^2\\ k& b^3& c^3& d^3\end{array}\right|=-k(b-c)(b-d)(c-d)$

${\Delta }_2=\left|\begin{array}{cccc}1& 0& 1& 1\\ a& 0& c& d\\ a^2& 0& c^2& d^2\\ a^3& k& c^3& d^3\end{array}\right|=-k(a-c)(a-d)(c-d)$

${\Delta }_3=\left|\begin{array}{cccc}1& 1& 0& 1\\ a& b& 0& d\\ a^2& b^2& 0& d^2\\ a^3& b^3& k& d^3\end{array}\right|=-k(a-b)(a-d)(b-d)$

${\Delta }_4=\left|\begin{array}{cccc}1& 1& 1& 0\\ a& b& c& 0\\ a^2& b^2& c^2& 0\\ a^3& b^3& c^3& k\end{array}\right|=-k(a-b)(a-c)(b-c)$

$x=\frac{{\Delta }_1}{\Delta }=-\frac{k}{(a-b)(a-c)(a-d)};y=\frac{{\Delta }_2}{\Delta }=-\frac{k}{(a-b)(b-c)(b-d)};z=\frac{{\Delta }_3}{\Delta }=-\frac{k}{(a-c)(b-c)(c-d)};u=\frac{{\Delta }_4}{\Delta }=-\frac{k}{(a-d)(b-d)(c-d)}$