Question

If $x,y,z\in R^{+}$ and $a,b,c$ are non-zero such that
$\left|\begin{array}{ccc}ax& cy& bz\\ by& az& cx\\ cz& bx& ay\end{array}\right|=\left|\begin{array}{ccc}ax& bx& cx\\ cy& ay& by\\ bz& cz& az\end{array}\right|,$
then which of the following is/are correct?

• A. $x+y+z=0$
• B. $x^2+y^2+z^2=xy+yz+zx$
• C. $x^3+y^3+z^3=3xyz$
• D. $x^4+y^4+z^4=2xyz(x+y+z)-(x^2{y}^2+y^2{z}^2+x^2{z}^2)$

Answer 1

Answer: (B), (C), (D)

$x^4+y^4+z^4=2xyz(x+y+z)-(x^2{y}^2+y^2{z}^2+x^2{z}^2)$

${\Delta }_1=\left|\begin{array}{ccc}ax& by& cz\\ cy& az& bx\\ bz& cx& ay\end{array}\right|[R\leftrightarrow C]\Rightarrow R_1\to yz{R}_1,R_2\to xz{R}_2,R_3\to xy{R}_3=\frac{1}{x^2{y}^2{z}^2}\left|\begin{array}{ccc}axyz& b{y}^{2}z& cy{z}^2\\ cxyz& a{z}^{2}x& b{x}^{2}z\\ bxyz& c{x}^{2}y& ax{y}^2\end{array}\right|=\frac{1}{xyz}\left|\begin{array}{ccc}a& b{y}^{2}z& cy{z}^2\\ c& a{z}^{2}x& b{x}^{2}z\\ b& c{x}^{2}y& ax{y}^2\end{array}\right|=\frac{1}{xyz}\left[x^2{y}^2{z}^2\right(a^3+b^3+c^3)-xyzabc(x^3+y^3+z^3\left)\right]=xyz(a^3+b^3+c^3)-abc(x^3+y^3+z^3)$

${\Delta }_2=xyz\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|=xyz(a^3+b^3+c^3-3abc)$

${\Delta }_1={\Delta }_{2}xyz(a^3+b^3+c^3)-abc(x^3+y^3+z^3)=xyz(a^3+b^3+c^3-3abc)\Rightarrow x^3+y^3+z^3-3xyz=0...\left(1\right)\Rightarrow x+y+z=0\text{or}x=y=z$
Since $x,y,z\in R^{+}$,
$x=y=z$
Hence option b, c, d are correct.