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Question Number 12261 by frank ntulah last updated on 17/Apr/17

find the value of a b and c a+b+c=4 a^2 +b^2 +c^2 =66 a^3 +b^3 +c^3 =280

$${find}\:{the}\:{value}\:{of}\:{a}\:{b}\:{and}\:{c} \\ $$$$\:{a}+{b}+{c}=\mathrm{4} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{66} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{280} \\ $$

Answered by mrW1 last updated on 17/Apr/17

(a+b+c)^2 =a^2 +b^2 +c^2 +2ab+2bc+2ca=4^2 =16 ⇒ab+bc+ca=((16−66)/2)=−25 (a+b+c)^3 =a^3 +b^3 +c^3 +6abc+3(a^2 b+a^2 c+b^2 a+b^2 c+c^2 a+c^2 b) =a^3 +b^3 +c^3 +3(a+b+c)(ab+bc+ca)−3abc =280+3×4×(−25)−3abc=4^3 =64 ⇒abc=((280+3×4×(−25)−64)/3)=−28 a,b,c are the roots of following cubic eqn. x^3 −(a+b+c)x^2 +(ab+bc+ca)x−abc=0 ⇒x^3 −4x^2 −25x+28=0 ⇒(x−1)(x+4)(x−7)=0 ⇒x_1 =−4, x_2 =1, x_3 =7 possible solutions for a,b,c are: ((a),(b),(c) )= (((−4)),(1),(7) ) ((a),(b),(c) )= (((−4)),(7),(1) ) ((a),(b),(c) )= ((1),((−4)),(7) ) ((a),(b),(c) )= ((1),(7),((−4)) ) ((a),(b),(c) )= ((7),((−4)),(1) ) ((a),(b),(c) )= ((7),(1),((−4)) )

$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ca}=\mathrm{4}^{\mathrm{2}} =\mathrm{16} \\ $$$$\Rightarrow{ab}+{bc}+{ca}=\frac{\mathrm{16}−\mathrm{66}}{\mathrm{2}}=−\mathrm{25} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\mathrm{6}{abc}+\mathrm{3}\left({a}^{\mathrm{2}} {b}+{a}^{\mathrm{2}} {c}+{b}^{\mathrm{2}} {a}+{b}^{\mathrm{2}} {c}+{c}^{\mathrm{2}} {a}+{c}^{\mathrm{2}} {b}\right) \\ $$$$={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\mathrm{3}\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)−\mathrm{3}{abc} \\ $$$$=\mathrm{280}+\mathrm{3}×\mathrm{4}×\left(−\mathrm{25}\right)−\mathrm{3}{abc}=\mathrm{4}^{\mathrm{3}} =\mathrm{64} \\ $$$$\Rightarrow{abc}=\frac{\mathrm{280}+\mathrm{3}×\mathrm{4}×\left(−\mathrm{25}\right)−\mathrm{64}}{\mathrm{3}}=−\mathrm{28} \\ $$$$ \\ $$$${a},{b},{c}\:{are}\:{the}\:{roots}\:{of}\:{following}\:{cubic}\:{eqn}. \\ $$$${x}^{\mathrm{3}} −\left({a}+{b}+{c}\right){x}^{\mathrm{2}} +\left({ab}+{bc}+{ca}\right){x}−{abc}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{25}{x}+\mathrm{28}=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{1}\right)\left({x}+\mathrm{4}\right)\left({x}−\mathrm{7}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}_{\mathrm{1}} =−\mathrm{4},\:{x}_{\mathrm{2}} =\mathrm{1},\:{x}_{\mathrm{3}} =\mathrm{7} \\ $$$${possible}\:{solutions}\:{for}\:{a},{b},{c}\:{are}: \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}=\begin{pmatrix}{−\mathrm{4}}\\{\mathrm{1}}\\{\mathrm{7}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}=\begin{pmatrix}{−\mathrm{4}}\\{\mathrm{7}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}}\\{−\mathrm{4}}\\{\mathrm{7}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{7}}\\{−\mathrm{4}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}=\begin{pmatrix}{\mathrm{7}}\\{−\mathrm{4}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}=\begin{pmatrix}{\mathrm{7}}\\{\mathrm{1}}\\{−\mathrm{4}}\end{pmatrix} \\ $$

Commented by frank ntulah last updated on 17/Apr/17

thank you sir you make me happy

$${thank}\:{you}\:{sir}\:{you}\:{make}\:{me}\:{happy} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by tawa last updated on 17/Apr/17

Yes sir, check line 4 ...... a^3 + b^3 + c^3 .....

$$\mathrm{Yes}\:\mathrm{sir},\:\mathrm{check}\:\mathrm{line}\:\mathrm{4}\:......\:\:\:\:\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \:+\:\mathrm{c}^{\mathrm{3}} \:..... \\ $$

Commented by mrW1 last updated on 17/Apr/17

Thanks. You are right. I have corrected.

$${Thanks}.\:{You}\:{are}\:{right}.\:{I}\:{have}\:{corrected}. \\ $$

Commented by tawa last updated on 17/Apr/17

Yes sir. Thank you. i always follow your steps very well. God bless you sir.

$$\mathrm{Yes}\:\mathrm{sir}.\:\mathrm{Thank}\:\mathrm{you}.\:\mathrm{i}\:\mathrm{always}\:\mathrm{follow}\:\mathrm{your}\:\mathrm{steps}\:\mathrm{very}\:\mathrm{well}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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