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Question-173629

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Question Number 173629 by mnjuly1970 last updated on 15/Jul/22
Commented by mnjuly1970 last updated on 15/Jul/22
solve for R a,b,c ? ⇑⇑⇑
$$\:\:\:\:{solve}\:\:{for}\:\:\:\mathbb{R}\:\:\:{a},{b},{c}\:?\:\Uparrow\Uparrow\Uparrow \\ $$
Commented by Tawa11 last updated on 15/Jul/22
Great sirs
$$\mathrm{Great}\:\mathrm{sirs} \\ $$
Answered by mahdipoor last updated on 15/Jul/22
a^3 +b^3 =(a+b)(a^2 +b^2 −ab)= a+b+(c)=a+b+(a+b)=2(a+b) i⇒a+b=0 ⇒ a=k , b=−k , c=0 (a^2 +b^2 =b+c) ⇒ 2k^2 =−k ⇒ k=0 or −(1/2) ⇒⇒ ((a),(b),(c) ) = ((0),(0),(0) ) or (((−0.5)),((0.5)),(0) ) ii⇒a+b≠0 ⇒ 2=(a^2 +b^2 )−ab=b+(c)−ab =2b+a−ab ⇒2b+a−ab−2=0 ⇒ (a−2)(1−b)=0 ⇒ 1)if a=2 ⇒ c=2+b & 4+b^2 =b+c ⇒ 4+b^2 =2b+2⇒b^2 −2b+2=0 ⇒ ∄b∈R 2)if b=1 ⇒ c=1+a & 1+a^2 =1+c ⇒ 1+a^2 =2+a⇒a^2 −a−1=0 ⇒a=((1±(√5))/2) ⇒⇒ ((a),(b),(c) ) = ((((1±(√5))/2)),(1),(((3±(√5))/2)) )
$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\left({a}+{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right)= \\ $$$${a}+{b}+\left({c}\right)={a}+{b}+\left({a}+{b}\right)=\mathrm{2}\left({a}+{b}\right) \\ $$$$\mathrm{i}\Rightarrow{a}+{b}=\mathrm{0}\:\Rightarrow\:{a}={k}\:,\:{b}=−{k}\:,\:{c}=\mathrm{0} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={b}+{c}\right)\:\Rightarrow\:\mathrm{2}{k}^{\mathrm{2}} =−{k}\:\Rightarrow\:{k}=\mathrm{0}\:{or}\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\Rightarrow\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{or}\:\begin{pmatrix}{−\mathrm{0}.\mathrm{5}}\\{\mathrm{0}.\mathrm{5}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{ii}\Rightarrow{a}+{b}\neq\mathrm{0}\:\Rightarrow\:\mathrm{2}=\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−{ab}={b}+\left({c}\right)−{ab} \\ $$$$=\mathrm{2}{b}+{a}−{ab}\:\Rightarrow\mathrm{2}{b}+{a}−{ab}−\mathrm{2}=\mathrm{0}\:\Rightarrow \\ $$$$\left({a}−\mathrm{2}\right)\left(\mathrm{1}−{b}\right)=\mathrm{0}\:\Rightarrow \\ $$$$\left.\mathrm{1}\right){if}\:{a}=\mathrm{2}\:\Rightarrow\:{c}=\mathrm{2}+{b}\:\&\:\mathrm{4}+{b}^{\mathrm{2}} ={b}+{c}\:\Rightarrow \\ $$$$\mathrm{4}+{b}^{\mathrm{2}} =\mathrm{2}{b}+\mathrm{2}\Rightarrow{b}^{\mathrm{2}} −\mathrm{2}{b}+\mathrm{2}=\mathrm{0}\:\Rightarrow\:\nexists{b}\in{R} \\ $$$$\left.\mathrm{2}\right){if}\:{b}=\mathrm{1}\:\Rightarrow\:{c}=\mathrm{1}+{a}\:\&\:\mathrm{1}+{a}^{\mathrm{2}} =\mathrm{1}+{c}\:\Rightarrow \\ $$$$\mathrm{1}+{a}^{\mathrm{2}} =\mathrm{2}+{a}\Rightarrow{a}^{\mathrm{2}} −{a}−\mathrm{1}=\mathrm{0}\:\Rightarrow{a}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\Rightarrow\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}\:=\begin{pmatrix}{\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}}\\{\mathrm{1}}\\{\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}}\end{pmatrix}\:\: \\ $$$$ \\ $$
Answered by MJS_new last updated on 15/Jul/22
obviously a=b=c=0 c=a+b b=pa { (((p^2 +1)a^2 −(2p+1)a=0)),(((p^3 +1)a^3 −2(p+1)a=0)) :} a≠0 { (((p^2 +1)a−2p−1=0)),(((p+1)((p^2 −p+1)a^2 −2)=0)) :} p_1 =−1 ⇒ a_1 =−(1/2)∧b_1 =(1/2)∧c_1 =0 a=((2p+1)/(p^2 +1)) p^4 −(3/2)p^2 +(3/2)p−(1/2)=0 (p^2 +p−1)(p^2 −p+(1/2))=0 p_2 =−(1/2)−((√5)/2) ⇒ a_2 =(1/2)−((√5)/2)∧b_2 =1∧c_2 =(3/2)−((√5)/2) p_3 =−(1/2)+((√5)/2) ⇒ a_3 =(1/2)+((√5)/2)∧b_3 =1∧c_3 =(3/2)+((√5)/2) (p_(4, 5) =(1/2)±(1/2)i)
$$\mathrm{obviously}\:{a}={b}={c}=\mathrm{0} \\ $$$${c}={a}+{b} \\ $$$${b}={pa} \\ $$$$\begin{cases}{\left({p}^{\mathrm{2}} +\mathrm{1}\right){a}^{\mathrm{2}} −\left(\mathrm{2}{p}+\mathrm{1}\right){a}=\mathrm{0}}\\{\left({p}^{\mathrm{3}} +\mathrm{1}\right){a}^{\mathrm{3}} −\mathrm{2}\left({p}+\mathrm{1}\right){a}=\mathrm{0}}\end{cases} \\ $$$${a}\neq\mathrm{0} \\ $$$$\begin{cases}{\left({p}^{\mathrm{2}} +\mathrm{1}\right){a}−\mathrm{2}{p}−\mathrm{1}=\mathrm{0}}\\{\left({p}+\mathrm{1}\right)\left(\left({p}^{\mathrm{2}} −{p}+\mathrm{1}\right){a}^{\mathrm{2}} −\mathrm{2}\right)=\mathrm{0}}\end{cases} \\ $$$${p}_{\mathrm{1}} =−\mathrm{1}\:\Rightarrow\:{a}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}\wedge{b}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\wedge{c}_{\mathrm{1}} =\mathrm{0} \\ $$$${a}=\frac{\mathrm{2}{p}+\mathrm{1}}{{p}^{\mathrm{2}} +\mathrm{1}} \\ $$$${p}^{\mathrm{4}} −\frac{\mathrm{3}}{\mathrm{2}}{p}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}{p}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\left({p}^{\mathrm{2}} +{p}−\mathrm{1}\right)\left({p}^{\mathrm{2}} −{p}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${p}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\:{a}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\wedge{b}_{\mathrm{2}} =\mathrm{1}\wedge{c}_{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${p}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\:{a}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\wedge{b}_{\mathrm{3}} =\mathrm{1}\wedge{c}_{\mathrm{3}} =\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left({p}_{\mathrm{4},\:\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}\right) \\ $$

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