Question Number 205626 by mr W last updated on 25/Mar/24
$${if}\:{a}+{b}+{c}=\frac{\mathrm{1}}{{a}+\mathrm{1}}+\frac{\mathrm{1}}{{b}+\mathrm{2}}+\frac{\mathrm{1}}{{c}+\mathrm{3}}=\mathrm{0}, \\ $$$${find}\:\left({a}+\mathrm{1}\right)^{\mathrm{2}} +\left({b}+\mathrm{2}\right)^{\mathrm{2}} +\left({c}+\mathrm{3}\right)^{\mathrm{2}} =? \\ $$
Answered by A5T last updated on 25/Mar/24
$${a}+{b}+{c}=\mathrm{0}\:\wedge\:\left({b}+\mathrm{2}\right)\left({c}+\mathrm{3}\right)+\left({a}+\mathrm{1}\right)\left({b}+\mathrm{2}\right)+\left({a}+\mathrm{1}\right)\left({c}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\left({a}+\mathrm{1}\right)^{\mathrm{2}} +\left({b}+\mathrm{2}\right)^{\mathrm{2}} +\left({c}+\mathrm{3}\right)^{\mathrm{2}} \\ $$$$=\left({a}+\mathrm{1}+{b}+\mathrm{2}+{c}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{2}\left({a}+\mathrm{1}\right)\left({b}+\mathrm{2}\right)−\mathrm{2}\left({a}+\mathrm{1}\right)\left({c}+\mathrm{3}\right) \\ $$$$−\mathrm{2}\left({c}+\mathrm{3}\right)\left({b}+\mathrm{2}\right)=\mathrm{6}^{\mathrm{2}} =\mathrm{36} \\ $$
Commented by mr W last updated on 26/Mar/24
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Answered by mr W last updated on 26/Mar/24
$${say}\:{x}={a}+\mathrm{1},\:{y}={b}+\mathrm{2},\:{z}={c}+\mathrm{3} \\ $$$$\Rightarrow{x}+{y}+{z}=\mathrm{1}+\mathrm{2}+\mathrm{3}=\mathrm{6} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\mathrm{0}\:\Rightarrow{xy}+{yz}+{zx}=\mathrm{0} \\ $$$$\left({a}+\mathrm{1}\right)^{\mathrm{2}} +\left({b}+\mathrm{2}\right)^{\mathrm{2}} +\left({c}+\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\:\:={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \\ $$$$\:\:=\left({x}+{y}+{z}\right)^{\mathrm{2}} −\mathrm{2}\left({xy}+{yz}+{zx}\right) \\ $$$$\:\:=\mathrm{6}^{\mathrm{2}} −\mathrm{2}×\mathrm{0}=\mathrm{36}\:\checkmark \\ $$