Question Number 36080 by Rio Mike last updated on 28/May/18
$$\:\mathrm{i}\:\mathrm{want}\:\mathrm{to}\:\mathrm{know}\:\mathrm{how}\: \\ $$$$\alpha^{\mathrm{2}} +\:\beta^{\mathrm{2}} =\:\left(\alpha+\beta\right)^{\mathrm{2}} −\:\mathrm{2}\alpha\beta\:\mathrm{why}\:\mathrm{not}\: \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\:\left(\alpha+\beta\right)^{\mathrm{2}} +\:\mathrm{2}\alpha\beta? \\ $$
Commented by Joel579 last updated on 29/May/18
$$\alpha^{\mathrm{2}} \:+\:\beta^{\mathrm{2}} \:=\:\left(\alpha\:+\:\beta\right)^{\mathrm{2}} \:−\:\mathrm{2}\alpha\beta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\alpha\:−\:\beta\right)^{\mathrm{2}} \:+\:\mathrm{2}\alpha\beta \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/May/18
$${value}\:{of}\:\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta \\ $$$$=\alpha^{\mathrm{2}} +\mathrm{2}\alpha\beta+\beta^{\mathrm{2}} −\mathrm{2}\alpha\beta \\ $$$$=\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \\ $$$${value}\:{of}\:\left(\alpha+\beta\right)^{\mathrm{2}} +\mathrm{2}\alpha\beta \\ $$$$=\alpha^{\mathrm{2}} +\mathrm{2}\alpha\beta+\beta^{\mathrm{2}} +\mathrm{2}\alpha\beta \\ $$$$=\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\mathrm{4}\alpha\beta \\ $$$${that}\:{is}\:{why}... \\ $$
Answered by Rasheed.Sindhi last updated on 28/May/18
$$\alpha^{\mathrm{2}} +\:\beta^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\alpha^{\mathrm{2}} +\mathrm{2}\alpha\beta+\:\beta^{\mathrm{2}} −\mathrm{2}\alpha\beta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta \\ $$