Question Number 204739 by hardmath last updated on 26/Feb/24
Answered by mr W last updated on 26/Feb/24
![x^4 +2x=−4 ⇒x^3 +2=−(4/x) ⇒(x^2 /(x^3 +2))=−(x^3 /4) Σ(a^2 /(a^3 +2))=−(1/4)Σa^3 =−(1/4)[0^3 −3×0×0+3×(−2)]=(3/2) ✓](https://www.tinkutara.com/question/Q204741.png)
$${x}^{\mathrm{4}} +\mathrm{2}{x}=−\mathrm{4} \\ $$$$\Rightarrow{x}^{\mathrm{3}} +\mathrm{2}=−\frac{\mathrm{4}}{{x}} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{3}} +\mathrm{2}}=−\frac{{x}^{\mathrm{3}} }{\mathrm{4}} \\ $$$$\Sigma\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{3}} +\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{4}}\Sigma{a}^{\mathrm{3}} \\ $$$$\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{0}^{\mathrm{3}} −\mathrm{3}×\mathrm{0}×\mathrm{0}+\mathrm{3}×\left(−\mathrm{2}\right)\right]=\frac{\mathrm{3}}{\mathrm{2}}\:\checkmark \\ $$
Commented by York12 last updated on 26/Feb/24
$$\mathrm{sir}\:\mathrm{may}\:\mathrm{you}\:\mathrm{elaborate}\:\mathrm{on}\:\mathrm{the}\:\mathrm{last}\:\mathrm{step} \\ $$
Commented by mr W last updated on 26/Feb/24
$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +{d}^{\mathrm{3}} =\left({a}+{b}+{c}+{d}\right)^{\mathrm{3}} −\mathrm{3}\left({a}+{b}+{c}+{d}\right)\left({ab}+{bc}+{cd}+{da}+{ac}+{bd}\right)+\mathrm{3}\left({abc}+{bcd}+{cda}+{abd}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}^{\mathrm{3}} −\mathrm{3}×\mathrm{0}×\mathrm{0}+\mathrm{3}×\left(−\mathrm{2}\right)=−\mathrm{6} \\ $$$${you}\:{can}\:{get}\:{this}\:{by}\:{using}\:{newton}'{s}\: \\ $$$${identities}: \\ $$
Commented by mr W last updated on 26/Feb/24
Commented by York12 last updated on 26/Feb/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by justenspi last updated on 28/Feb/24
![x^4 +2x+4=0 ⇔x^3 +2=((−4)/x)⇔(1/(x^3 +2))=((−x)/4) ⇔(x^2 /(x^3 +2))=((−x^3 )/4) ⇒Σ_(cyc) ((a^2 /(a^3 +2)))=((−1)/4)Σ_(cyc) (a^3 ) ,According to vieta formulae we have : Σ_(cyc) (a)=0 ,Σ_(cyc) (ab)=0 ,Σ_(cyc) (abc)=−2,abcd=4 ∴ We have Σ_(cyc) (a^3 )=[Σ_(cyc) (a)].[Σ_(cyc) (a^2 )]−Σ_(cyc) (a(b^2 +c^2 +d^2 ))=−Σ_(cyc) (a((b+c+d)^2 −2(bc+cd+db)) =−Σ_(cyc) (a^3 )+2Σ_(cyc) (abc+acd+adb)=−Σ_(cyc) (a^3 )+6Σ_(cyc) (abc) ⇔Σ_(cyc) (a^3 )=3Σ_(cyc) (abc)=−6 ∴Σ_(cyc) ((a^2 /(a^3 +2)))=((−1)/4)Σ_(cyc) (a^3 )=(6/4)=(3/2) □](https://www.tinkutara.com/question/Q204824.png)
$$ \\ $$$${x}^{\mathrm{4}} +\mathrm{2}{x}+\mathrm{4}=\mathrm{0}\:\Leftrightarrow{x}^{\mathrm{3}} +\mathrm{2}=\frac{−\mathrm{4}}{{x}}\Leftrightarrow\frac{\mathrm{1}}{{x}^{\mathrm{3}} +\mathrm{2}}=\frac{−{x}}{\mathrm{4}}\:\Leftrightarrow\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{3}} +\mathrm{2}}=\frac{−{x}^{\mathrm{3}} }{\mathrm{4}} \\ $$$$\Rightarrow\underset{{cyc}} {\sum}\left(\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{3}} +\mathrm{2}}\right)=\frac{−\mathrm{1}}{\mathrm{4}}\underset{{cyc}} {\sum}\left({a}^{\mathrm{3}} \right)\:,\mathrm{According}\:\mathrm{to}\:\mathrm{vieta}\:\mathrm{formulae}\:\mathrm{we}\:\mathrm{have}\:: \\ $$$$\underset{{cyc}} {\sum}\left({a}\right)=\mathrm{0}\:,\underset{{cyc}} {\sum}\left({ab}\right)=\mathrm{0}\:,\underset{{cyc}} {\sum}\left({abc}\right)=−\mathrm{2},{abcd}=\mathrm{4}\: \\ $$$$\therefore\:\mathrm{We}\:\mathrm{have}\:\underset{{cyc}} {\sum}\left({a}^{\mathrm{3}} \right)=\left[\underset{{cyc}} {\sum}\left({a}\right)\right].\left[\underset{{cyc}} {\sum}\left({a}^{\mathrm{2}} \right)\right]−\underset{{cyc}} {\sum}\left({a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\right)=−\underset{{cyc}} {\sum}\left({a}\left(\left({b}+{c}+{d}\right)^{\mathrm{2}} −\mathrm{2}\left({bc}+{cd}+{db}\right)\right)\right. \\ $$$$=−\underset{{cyc}} {\sum}\left({a}^{\mathrm{3}} \right)+\mathrm{2}\underset{{cyc}} {\sum}\left({abc}+{acd}+{adb}\right)=−\underset{{cyc}} {\sum}\left({a}^{\mathrm{3}} \right)+\mathrm{6}\underset{{cyc}} {\sum}\left({abc}\right)\:\Leftrightarrow\underset{{cyc}} {\sum}\left({a}^{\mathrm{3}} \right)=\mathrm{3}\underset{{cyc}} {\sum}\left({abc}\right)=−\mathrm{6} \\ $$$$\therefore\underset{{cyc}} {\sum}\left(\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{3}} +\mathrm{2}}\right)=\frac{−\mathrm{1}}{\mathrm{4}}\underset{{cyc}} {\sum}\left({a}^{\mathrm{3}} \right)=\frac{\mathrm{6}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{2}}\:\Box \\ $$