Question Number 26250 by Chuks” last updated on 23/Dec/17
$${someone}\:{should}\:{help}\:{witb}\:{solution}\:{please} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{3}{y}+\mathrm{4} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} −\mathrm{6}{x}+{y}−\mathrm{10}=\sqrt{\left({y}+\mathrm{5}\right)}−\sqrt{\left(\mathrm{4}{x}+{y}\right)} \\ $$
Answered by jota@ last updated on 25/Dec/17
![If x+y=1 then x^3 +y^3 +3xy(x+y)=1 x^3 +y^3 +3xy=1 x^3 +y^3 +3x(1−x)=1 x^3 +y^3 =3x^2 −3x+1+[−3(x+y)+3] x^3 +y^3 =3x^2 −6x−3y+4. Continue](https://www.tinkutara.com/question/Q26394.png)
$$ \\ $$$${If}\:\:{x}+{y}=\mathrm{1}\:\:{then} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +\mathrm{3}{xy}\left({x}+{y}\right)=\mathrm{1}\: \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +\mathrm{3}{xy}=\mathrm{1}\: \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +\mathrm{3}{x}\left(\mathrm{1}−{x}\right)=\mathrm{1}\: \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}+\left[−\mathrm{3}\left({x}+{y}\right)+\mathrm{3}\right] \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{3}{y}+\mathrm{4}. \\ $$$${Continue} \\ $$$$ \\ $$
Commented by Chuks” last updated on 29/Dec/17
$${please}\:{help}\:{me}\:{finih}\:{it}…{please}\:{Prof}\:{m} \\ $$