Menu Close

Let-x-y-z-be-real-numbers-satisfying-the-equations-x-y-z-7-xy-yz-zx-10-xyz-6-Find-the-value-of-x-3-y-3-z-3-

Tinku Tara 0 Comments
Pin




Question Number 217377 by ArshadS last updated on 12/Mar/25
Let x,y,z be real numbers satisfying the equations x + y + z= 7 xy + yz + zx=10 xyz=6 Find the value of x^3 + y^3 + z^3
$$\mathrm{Let}\:\mathrm{x},\mathrm{y},\mathrm{z}\:\mathrm{be}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{equations} \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}=\:\mathrm{7} \\ $$$$\mathrm{xy}\:+\:\mathrm{yz}\:+\:\mathrm{zx}=\mathrm{10} \\ $$$$\mathrm{xyz}=\mathrm{6} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} \\ $$
Answered by Rasheed.Sindhi last updated on 12/Mar/25
{ ((x + y + z= 7...(i))),((xy + yz + zx=10...(ii))),((xyz=6...(iii))),((x^3 + y^3 + z^3 =?)) :} (i)^2 : x^2 +y^2 +z^2 +2(xy+yz+zx)=49 x^2 +y^2 +z^2 +2(10)=49 x^2 +y^2 +z^2 =29 x^3 + y^3 + z^3 −3xyz=(x+y+z)(x^2 +y^2 +z^2 −(xy+yz+zx) ) x^3 + y^3 + z^3 −3(6)=(7)(29−(10) ) x^3 + y^3 + z^3 =18+(7)(29−(10) ) =18+133=151
$$\begin{cases}{\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}=\:\mathrm{7}…\left({i}\right)}\\{\mathrm{xy}\:+\:\mathrm{yz}\:+\:\mathrm{zx}=\mathrm{10}…\left({ii}\right)}\\{\mathrm{xyz}=\mathrm{6}…\left({iii}\right)}\\{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} =?}\end{cases} \\ $$$$\left({i}\right)^{\mathrm{2}} :\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)=\mathrm{49} \\ $$$$\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{10}\right)=\mathrm{49} \\ $$$$\:\:\:\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{29} \\ $$$$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} −\mathrm{3xyz}=\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} −\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)\:\right) \\ $$$$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} −\mathrm{3}\left(\mathrm{6}\right)=\left(\mathrm{7}\right)\left(\mathrm{29}−\left(\mathrm{10}\right)\:\right) \\ $$$$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} =\mathrm{18}+\left(\mathrm{7}\right)\left(\mathrm{29}−\left(\mathrm{10}\right)\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{18}+\mathrm{133}=\mathrm{151} \\ $$
Commented by ArshadS last updated on 13/Mar/25
Thanks sir!
$${Thanks}\:{sir}! \\ $$
Answered by mr W last updated on 12/Mar/25
(x+y+z)^3 =x^3 +y^3 +z^3 −3xyz+3(x+y+z)(xy+yz+zx) 7^3 =x^3 +y^3 +z^3 −3×6+3×7×10 ⇒x^3 +y^3 +z^3 =7^3 +3×6−3×7×10=151
$$\left({x}+{y}+{z}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} −\mathrm{3}{xyz}+\mathrm{3}\left({x}+{y}+{z}\right)\left({xy}+{yz}+{zx}\right) \\ $$$$\mathrm{7}^{\mathrm{3}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} −\mathrm{3}×\mathrm{6}+\mathrm{3}×\mathrm{7}×\mathrm{10} \\ $$$$\Rightarrow{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{7}^{\mathrm{3}} +\mathrm{3}×\mathrm{6}−\mathrm{3}×\mathrm{7}×\mathrm{10}=\mathrm{151} \\ $$
Commented by ArshadS last updated on 13/Mar/25
Thanks sir!
$$\mathrm{Thanks}\:\mathrm{sir}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *