Question Number 172014 by Mikenice last updated on 23/Jun/22
$${using}\:{properties}\:{of}\:{determinats} \\ $$$${prove}\:{that} \\ $$$$\left[−{yz}\:\:\:\:\:\:{y}^{\mathrm{2}} +{yz}\:\:\:\:\:\:\:{z}^{\mathrm{2}} +{yz}\right] \\ $$$$\left[{x}^{\mathrm{2}} +{xz}\:\:\:−{xz}\:\:\:\:\:\:\:\:\:\:{z}^{\mathrm{2}} +{xy}\right]\:=\left({xy}+{yz}+{zx}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\left.\:{x}^{\mathrm{2}} +{xy}\:\:\:\:\:{y}^{\mathrm{2}} +{xy}\:\:\:\:\:\:\:\:−{xy}\right] \\ $$
Commented by Mikenice last updated on 23/Jun/22
$${okay}\:{sir} \\ $$
Commented by som(math1967) last updated on 23/Jun/22
$${I}\:{think}\:{question}\:{is}\:\boldsymbol{{wrong}} \\ $$