Question Number 49642 by maxmathsup by imad last updated on 08/Dec/18
$${if}\:\:{a}+{b}\:={s}\:{and}\:{a}^{\mathrm{3}} \:+{b}^{\mathrm{3}} \:={t}\:\:{find}\:{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \:\:{and}\:{a}^{\mathrm{4}} \:+{b}^{\mathrm{4}} \:{interms}\:{of}\:{s}\:{and}\:{t}\:. \\ $$
Answered by mr W last updated on 08/Dec/18
![a^3 +b^3 =(a+b)(a^2 −ab+b^2 )=(a+b)[(a+b)^2 −3ab] t=s(s^2 −3ab) ⇒ab=(1/3)(s^2 −(t/s)) a^2 +b^2 =a^2 +2ab+b^2 −2ab=(a+b)^2 −2ab =s^2 −(2/3)(s^2 −(t/s))=(1/3)(s^2 +((2t)/s)) a^4 +b^4 =(a^2 +b^2 )^2 −2(ab)^2 =(1/9)(s^2 +((2t)/s))^2 −(2/9)(s^2 −(t/s))^2 =(1/9)(s^4 +((4t^2 )/s^2 )+2st)−(2/9)(s^4 +(t^2 /s^2 )−2st) =(1/9)(((2t^2 )/s^2 )+6st−s^4 )](https://www.tinkutara.com/question/Q49643.png)
$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right)=\left({a}+{b}\right)\left[\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{3}{ab}\right] \\ $$$${t}={s}\left({s}^{\mathrm{2}} −\mathrm{3}{ab}\right) \\ $$$$\Rightarrow{ab}=\frac{\mathrm{1}}{\mathrm{3}}\left({s}^{\mathrm{2}} −\frac{{t}}{{s}}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} −\mathrm{2}{ab}=\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab} \\ $$$$={s}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}\left({s}^{\mathrm{2}} −\frac{{t}}{{s}}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left({s}^{\mathrm{2}} +\frac{\mathrm{2}{t}}{{s}}\right) \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} =\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({ab}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{9}}\left({s}^{\mathrm{2}} +\frac{\mathrm{2}{t}}{{s}}\right)^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{9}}\left({s}^{\mathrm{2}} −\frac{{t}}{{s}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left({s}^{\mathrm{4}} +\frac{\mathrm{4}{t}^{\mathrm{2}} }{{s}^{\mathrm{2}} }+\mathrm{2}{st}\right)−\frac{\mathrm{2}}{\mathrm{9}}\left({s}^{\mathrm{4}} +\frac{{t}^{\mathrm{2}} }{{s}^{\mathrm{2}} }−\mathrm{2}{st}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left(\frac{\mathrm{2}{t}^{\mathrm{2}} }{{s}^{\mathrm{2}} }+\mathrm{6}{st}−{s}^{\mathrm{4}} \right) \\ $$
Commented by maxmathsup by imad last updated on 08/Dec/18
$${thank}\:{you}\:{sir}\:. \\ $$