Question

Solve the following equations:
$x^2+y^2+z^2=yz+zx+xy=a^2,3x-y+z=a\sqrt{3}$.

Answer 1

$x^2+y^2+z^2=xy+yz+zx=a^2$ ......... (i)

$3x-y+z=a\sqrt{3}$ ........... (ii)

$⟹2(x^2+y^2+z^2)=2a^2$ ....... (iii)

$2(xy+yz+zx)=2a^2$ ........ (iv) From (i)
Subtracting $\left(iii\right)$ and $\left(iv\right):-$

$2(x^2+y^2+z^2)-2xy-2yz-2zx=0$

$x^2+y^2-2xy+y^2+z^2-2yz+z^2+x^2-2zx=0$

${(x-y)}^2+{(y-z)}^2+{(z-x)}^2=0$

This is possible iff,
$x-y=0,y-z=0,z-x=0$

i.e. $x=y,y=z,z=x$

i.e. $x=y=z=k$ (let)

In equation $\left(ii\right)$

$3k-k+k=a\sqrt{3}$

$\Rightarrow k=\frac{a}{\sqrt{3}}$

Hence, $x=y=z=\frac{a}{\sqrt{3}}$