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x-y-z-7-x-2-y-2-z-2-49-x-3-y-3-z-3-7-find-x-y-z-




Question Number 95106 by bobhans last updated on 23/May/20
 { ((x+y+z = 7)),((x^2 +y^2 +z^2  = 49)),((x^3 +y^3 +z^3  = 7)) :}  find x; y ; z
$$\begin{cases}{{x}+{y}+{z}\:=\:\mathrm{7}}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \:=\:\mathrm{49}}\\{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} \:=\:\mathrm{7}}\end{cases} \\ $$$${find}\:{x};\:\mathrm{y}\:;\:{z}\: \\ $$
Answered by john santu last updated on 23/May/20
(1)x^3 +y^3 +z^3 −3xyz = (x+y+z)(x^2 +y^2 +z^2 −(xy+xz+yz))  ⇒ 7−3xyz = 7(49−(xy+xz+yz))  (2) (x+y+z)^2 = x^2 +y^2 +z^2 +2(xy+xz+yz)  ⇒49 = 49 +2(xy+xz+yz)  xy+xz+yz = 0  so we get 7−3xyz = 7×49  xyz = − 112
$$\left(\mathrm{1}\right){x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} −\mathrm{3}{xyz}\:=\:\left({x}+{y}+{z}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\left({xy}+{xz}+{yz}\right)\right) \\ $$$$\Rightarrow\:\mathrm{7}−\mathrm{3}{xyz}\:=\:\mathrm{7}\left(\mathrm{49}−\left({xy}+{xz}+{yz}\right)\right) \\ $$$$\left(\mathrm{2}\right)\:\left({x}+{y}+{z}\right)^{\mathrm{2}} =\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}\left({xy}+{xz}+{yz}\right) \\ $$$$\Rightarrow\mathrm{49}\:=\:\mathrm{49}\:+\mathrm{2}\left({xy}+{xz}+{yz}\right) \\ $$$${xy}+{xz}+{yz}\:=\:\mathrm{0} \\ $$$${so}\:{we}\:{get}\:\mathrm{7}−\mathrm{3}{xyz}\:=\:\mathrm{7}×\mathrm{49} \\ $$$${xyz}\:=\:−\:\mathrm{112}\: \\ $$
Commented by 1549442205 last updated on 23/May/20
That is not fully answer!I have the following answer:  We have 2(xy+yz+zx)=(x+y+z)^2 −(x^2 +y^2   +z^2 )=7^2 −49=0→xy+yz+zx=0→yz+  (y+z)x=0→yz+(7−x)x=0→yz=x^2 −7x(1)  On the other hands,x^3 +y^3 +z^3 =7⇔  x^3 +[(y+z)^3 −3yz(y+z)]=7⇔x^3 +[(7−x)^3 −  3(x^2 −7x)(7−x)]=7⇔x^3 +[343−147x+  21x^2 −x^3 −3(−x^3 +14x^2 −49x)]=7⇔  3x^3 −21x^2 +336=0⇔x^3 −7x^2 +112=0.  Putting x=(1/t) we get  112t^3 −7t+1=0.Multiplying two sides of  the equation by 196=7^2 .4 and note 112=  7.4^2  then set u=28t we get PT: u^3 −49u+  196=0(1)  Put u=−(m+n)→u+m+n=0.Using the  equality a^3 +b^3 +c^3 −3abc=(a+b+c)(a^2 +  b^2 +c^2 −ab−bc−ca) we have u^3 +m^3 +n^3 −  3umn=0(2).From (1) and (2) we have:  m^3 +n^3 =196(3),mn=((49)/3).It follows   (m^3 −n^3 )^2 =(m^3 +n^3 )^2 −4(mn)^3 =196^2 −  4.(((49)/3))^3 =(((98)/9)(√(177)))^2 ⇒m^3 −n^3 =((98)/9)(√(177))(2)  .Hence,from (1),(2)we get m^3 =(  ((98(√(177)))/9)+196)/2 and n^3 =(196−((98(√(177)))/9))/2  ⇒u=−(m+n)=  −[(^3 (√((1764+98(√(177)))/(18))) +^3 (√((1764−98(√(177)))/(18))))  Since u=28tand t=(1/x)⇒x=((28)/u)=−(1/7)[^3 (√((((1764+98(√(177)))/(18)))^2 ))−((49)/3)+^3 (√((((1764−98(√(177)))/(18)))^2 ))]  Replace this value of x into the equations:  x+y+z=7 and x(y+z)+yz=0 we find y,z   which are image numbers
$$\mathrm{That}\:\mathrm{is}\:\mathrm{not}\:\mathrm{fully}\:\mathrm{answer}!\mathrm{I}\:\mathrm{have}\:\mathrm{the}\:\mathrm{following}\:\mathrm{answer}: \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{2}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)=\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)^{\mathrm{2}} −\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right. \\ $$$$\left.+\mathrm{z}^{\mathrm{2}} \right)=\mathrm{7}^{\mathrm{2}} −\mathrm{49}=\mathrm{0}\rightarrow\mathrm{xy}+\mathrm{yz}+\mathrm{zx}=\mathrm{0}\rightarrow\mathrm{yz}+ \\ $$$$\left(\mathrm{y}+\mathrm{z}\right)\mathrm{x}=\mathrm{0}\rightarrow\mathrm{yz}+\left(\mathrm{7}−\mathrm{x}\right)\mathrm{x}=\mathrm{0}\rightarrow\mathrm{yz}=\mathrm{x}^{\mathrm{2}} −\mathrm{7x}\left(\mathrm{1}\right) \\ $$$$\mathrm{On}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hands},\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} +\mathrm{z}^{\mathrm{3}} =\mathrm{7}\Leftrightarrow \\ $$$$\mathrm{x}^{\mathrm{3}} +\left[\left(\mathrm{y}+\mathrm{z}\right)^{\mathrm{3}} −\mathrm{3yz}\left(\mathrm{y}+\mathrm{z}\right)\right]=\mathrm{7}\Leftrightarrow\mathrm{x}^{\mathrm{3}} +\left[\left(\mathrm{7}−\mathrm{x}\right)^{\mathrm{3}} −\right. \\ $$$$\left.\mathrm{3}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{7x}\right)\left(\mathrm{7}−\mathrm{x}\right)\right]=\mathrm{7}\Leftrightarrow\mathrm{x}^{\mathrm{3}} +\left[\mathrm{343}−\mathrm{147x}+\right. \\ $$$$\left.\mathrm{21x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{3}} −\mathrm{3}\left(−\mathrm{x}^{\mathrm{3}} +\mathrm{14x}^{\mathrm{2}} −\mathrm{49x}\right)\right]=\mathrm{7}\Leftrightarrow \\ $$$$\mathrm{3x}^{\mathrm{3}} −\mathrm{21x}^{\mathrm{2}} +\mathrm{336}=\mathrm{0}\Leftrightarrow\mathrm{x}^{\mathrm{3}} −\mathrm{7x}^{\mathrm{2}} +\mathrm{112}=\mathrm{0}. \\ $$$$\mathrm{Putting}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{112t}^{\mathrm{3}} −\mathrm{7t}+\mathrm{1}=\mathrm{0}.\mathrm{Multiplying}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{by}\:\mathrm{196}=\mathrm{7}^{\mathrm{2}} .\mathrm{4}\:\mathrm{and}\:\mathrm{note}\:\mathrm{112}= \\ $$$$\mathrm{7}.\mathrm{4}^{\mathrm{2}} \:\mathrm{then}\:\mathrm{set}\:\mathrm{u}=\mathrm{28t}\:\mathrm{we}\:\mathrm{get}\:\mathrm{PT}:\:\mathrm{u}^{\mathrm{3}} −\mathrm{49u}+ \\ $$$$\mathrm{196}=\mathrm{0}\left(\mathrm{1}\right) \\ $$$$\mathrm{Put}\:\mathrm{u}=−\left(\mathrm{m}+\mathrm{n}\right)\rightarrow\mathrm{u}+\mathrm{m}+\mathrm{n}=\mathrm{0}.\mathrm{Using}\:\mathrm{the} \\ $$$$\mathrm{equality}\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} −\mathrm{3abc}=\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{a}^{\mathrm{2}} +\right. \\ $$$$\left.\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{ab}−\mathrm{bc}−\mathrm{ca}\right)\:\mathrm{we}\:\mathrm{have}\:\mathrm{u}^{\mathrm{3}} +\mathrm{m}^{\mathrm{3}} +\mathrm{n}^{\mathrm{3}} − \\ $$$$\mathrm{3umn}=\mathrm{0}\left(\mathrm{2}\right).\mathrm{From}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right)\:\mathrm{we}\:\mathrm{have}: \\ $$$$\mathrm{m}^{\mathrm{3}} +\mathrm{n}^{\mathrm{3}} =\mathrm{196}\left(\mathrm{3}\right),\mathrm{mn}=\frac{\mathrm{49}}{\mathrm{3}}.\mathrm{It}\:\mathrm{follows}\: \\ $$$$\left(\mathrm{m}^{\mathrm{3}} −\mathrm{n}^{\mathrm{3}} \right)^{\mathrm{2}} =\left(\mathrm{m}^{\mathrm{3}} +\mathrm{n}^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{mn}\right)^{\mathrm{3}} =\mathrm{196}^{\mathrm{2}} − \\ $$$$\mathrm{4}.\left(\frac{\mathrm{49}}{\mathrm{3}}\right)^{\mathrm{3}} =\left(\frac{\mathrm{98}}{\mathrm{9}}\sqrt{\mathrm{177}}\right)^{\mathrm{2}} \Rightarrow\mathrm{m}^{\mathrm{3}} −\mathrm{n}^{\mathrm{3}} =\frac{\mathrm{98}}{\mathrm{9}}\sqrt{\mathrm{177}}\left(\mathrm{2}\right) \\ $$$$.\mathrm{Hence},\mathrm{from}\:\left(\mathrm{1}\right),\left(\mathrm{2}\right)\mathrm{we}\:\mathrm{get}\:\mathrm{m}^{\mathrm{3}} =\left(\right. \\ $$$$\left.\frac{\mathrm{98}\sqrt{\mathrm{177}}}{\mathrm{9}}+\mathrm{196}\right)/\mathrm{2}\:\mathrm{and}\:\mathrm{n}^{\mathrm{3}} =\left(\mathrm{196}−\frac{\mathrm{98}\sqrt{\mathrm{177}}}{\mathrm{9}}\right)/\mathrm{2} \\ $$$$\Rightarrow\mathrm{u}=−\left(\mathrm{m}+\mathrm{n}\right)= \\ $$$$−\left[\left(\:^{\mathrm{3}} \sqrt{\frac{\mathrm{1764}+\mathrm{98}\sqrt{\mathrm{177}}}{\mathrm{18}}}\:+^{\mathrm{3}} \sqrt{\frac{\mathrm{1764}−\mathrm{98}\sqrt{\mathrm{177}}}{\mathrm{18}}}\right)\right. \\ $$$$\mathrm{Since}\:\mathrm{u}=\mathrm{28tand}\:\mathrm{t}=\frac{\mathrm{1}}{\mathrm{x}}\Rightarrow\mathrm{x}=\frac{\mathrm{28}}{\mathrm{u}}=−\frac{\mathrm{1}}{\mathrm{7}}\left[^{\mathrm{3}} \sqrt{\left(\frac{\mathrm{1764}+\mathrm{98}\sqrt{\mathrm{177}}}{\mathrm{18}}\right)^{\mathrm{2}} }−\frac{\mathrm{49}}{\mathrm{3}}+^{\mathrm{3}} \sqrt{\left(\frac{\mathrm{1764}−\mathrm{98}\sqrt{\mathrm{177}}}{\mathrm{18}}\right)^{\mathrm{2}} }\right] \\ $$$$\mathrm{Replace}\:\mathrm{this}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{into}\:\mathrm{the}\:\mathrm{equations}: \\ $$$$\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{7}\:\mathrm{and}\:\mathrm{x}\left(\mathrm{y}+\mathrm{z}\right)+\mathrm{yz}=\mathrm{0}\:\mathrm{we}\:\mathrm{find}\:\mathrm{y},\mathrm{z}\: \\ $$$$\mathrm{which}\:\mathrm{are}\:\mathrm{image}\:\mathrm{numbers} \\ $$$$ \\ $$
Commented by john santu last updated on 23/May/20
what your answer?
$$\mathrm{what}\:\mathrm{your}\:\mathrm{answer}? \\ $$
Commented by bobhans last updated on 23/May/20
waw....superb sir
$$\mathrm{waw}….\mathrm{superb}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 23/May/20
e_1 =p_1 =x+y+z=7  2e_2 =p_1 ^2 −p_2 =7^2 −49=0 ⇒xy+yz+zx=0  3e_3 =(1/2)p_1 ^3 −(3/2)p_1 p_2 +p_3 =(7^3 /2)−((3×7×49)/2)+7=−336 ⇒xyz=−112  x,y,z are roots of  t^3 −7t^2 +112=0  ...
$${e}_{\mathrm{1}} ={p}_{\mathrm{1}} ={x}+{y}+{z}=\mathrm{7} \\ $$$$\mathrm{2}{e}_{\mathrm{2}} ={p}_{\mathrm{1}} ^{\mathrm{2}} −{p}_{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} −\mathrm{49}=\mathrm{0}\:\Rightarrow{xy}+{yz}+{zx}=\mathrm{0} \\ $$$$\mathrm{3}{e}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}{p}_{\mathrm{1}} ^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}}{p}_{\mathrm{1}} {p}_{\mathrm{2}} +{p}_{\mathrm{3}} =\frac{\mathrm{7}^{\mathrm{3}} }{\mathrm{2}}−\frac{\mathrm{3}×\mathrm{7}×\mathrm{49}}{\mathrm{2}}+\mathrm{7}=−\mathrm{336}\:\Rightarrow{xyz}=−\mathrm{112} \\ $$$${x},{y},{z}\:{are}\:{roots}\:{of} \\ $$$${t}^{\mathrm{3}} −\mathrm{7}{t}^{\mathrm{2}} +\mathrm{112}=\mathrm{0} \\ $$$$… \\ $$
Commented by bobhans last updated on 23/May/20
alright, the answer is very ′smart′
$$\mathrm{alright},\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{very}\:'\mathrm{smart}'\: \\ $$
Commented by bobhans last updated on 23/May/20
not finish ?
$$\mathrm{not}\:\mathrm{finish}\:? \\ $$
Commented by mr W last updated on 23/May/20
the rest is just “stupid” calculation   with values. you can do it when you  like.
$${the}\:{rest}\:{is}\:{just}\:“{stupid}''\:{calculation}\: \\ $$$${with}\:{values}.\:{you}\:{can}\:{do}\:{it}\:{when}\:{you} \\ $$$${like}. \\ $$
Commented by mr W last updated on 23/May/20
i think the main point of this question  is to get that x,y,z are the three roots  of eqn. t^3 −7t^2 +112=0. to calculate  the roots (one real and two complex)  of the cubic eqn.  is a hard but also easy  diligence work.
$${i}\:{think}\:{the}\:{main}\:{point}\:{of}\:{this}\:{question} \\ $$$${is}\:{to}\:{get}\:{that}\:{x},{y},{z}\:{are}\:{the}\:{three}\:{roots} \\ $$$${of}\:{eqn}.\:{t}^{\mathrm{3}} −\mathrm{7}{t}^{\mathrm{2}} +\mathrm{112}=\mathrm{0}.\:{to}\:{calculate} \\ $$$${the}\:{roots}\:\left({one}\:{real}\:{and}\:{two}\:{complex}\right) \\ $$$${of}\:{the}\:{cubic}\:{eqn}.\:\:{is}\:{a}\:{hard}\:{but}\:{also}\:{easy} \\ $$$${diligence}\:{work}. \\ $$
Commented by 1549442205 last updated on 23/May/20
This equation has unique real root   t=((−1)/7)[^3 (√((((1764+98(√(177)))/(18)))^2 ))−((49)/3)+^3 (√((((1764−98(√(177)))/(18)))^2 ))]like I done above.
$$\mathrm{This}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{unique}\:\mathrm{real}\:\mathrm{root}\: \\ $$$$\mathrm{t}=\frac{−\mathrm{1}}{\mathrm{7}}\left[^{\mathrm{3}} \sqrt{\left(\frac{\mathrm{1764}+\mathrm{98}\sqrt{\mathrm{177}}}{\mathrm{18}}\right)^{\mathrm{2}} }−\frac{\mathrm{49}}{\mathrm{3}}+^{\mathrm{3}} \sqrt{\left(\frac{\mathrm{1764}−\mathrm{98}\sqrt{\mathrm{177}}}{\mathrm{18}}\right)^{\mathrm{2}} }\right]\mathrm{like}\:\mathrm{I}\:\mathrm{done}\:\mathrm{above}. \\ $$

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