Question Number 98135 by Tony Lin last updated on 11/Jun/20
$${how}\:{to}\:{split}\:{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \:{into} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({x}+{y}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{y}\right)^{\mathrm{2}} \:? \\ $$
Commented by MJS last updated on 11/Jun/20
$${x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \neq\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({x}+{y}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{y}\right)^{\mathrm{2}} \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{something}\:\mathrm{wrong}\:\mathrm{with}\:\mathrm{this}\:\mathrm{question} \\ $$
Commented by MJS last updated on 11/Jun/20
$${x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} ={p}\left({x}+{y}\right)^{\mathrm{2}} +{q}\left({x}−{y}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow \\ $$$$\left(\mathrm{1}−{p}−{q}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\left(\mathrm{1}−\mathrm{2}{p}+\mathrm{2}{q}\right){xy}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{1}−{p}−{q}=\mathrm{0}\wedge\mathrm{1}−\mathrm{2}{p}+\mathrm{2}{q}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${p}=\frac{\mathrm{3}}{\mathrm{4}}\wedge{q}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow \\ $$$$\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({x}+{y}\right)\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{y}\right)\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \\ $$