Question Number 173078 by AgniMath last updated on 06/Jul/22
Answered by som(math1967) last updated on 07/Jul/22
![let b−c=x ,(a−b)=y,c−a=z ∴ x+y+z=0 (x+y)^2 =(−z)^2 x^2 +y^2 −z^2 =−2xy ⇒(x^2 +y^2 −z^2 )^2 =4x^2 y^2 ⇒x^4 +y^4 +z^4 =4x^2 y^2 −2x^2 y^2 +2y^2 z^2 +2z^2 x^2 ⇒x^4 +y^4 +z^4 =2(x^2 y^2 +y^2 z^2 +z^2 x^2 ) ⇒(1/2)(x^4 +y^4 +z^4 )=(xy)^2 +(yz)^2 +(zx)^2 +2xyz(x+y+z) [∵ x+y+z=0] (1/2)(x^4 +y^4 +z^4 )=(xy)^2 +(yz)^2 +(zx)^2 2xy.yz+2xzyz+2xy.zx ⇒(1/2)(x^4 +y^4 +z^4 )=(xy+yz+zx)^2 ∴(1/2){(b−c)^4 +(c−a)^4 +(a−b)^4 }= {(b−c)(a−b)+(c−a)(a−b)+(b−c)(c−a)}^2](https://www.tinkutara.com/question/Q173131.png)
$$\boldsymbol{{let}}\:\boldsymbol{{b}}−\boldsymbol{{c}}=\boldsymbol{{x}}\:,\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)=\boldsymbol{{y}},\boldsymbol{{c}}−\boldsymbol{{a}}=\boldsymbol{{z}} \\ $$$$\therefore\:\boldsymbol{{x}}+\boldsymbol{{y}}+\boldsymbol{{z}}=\mathrm{0} \\ $$$$\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right)^{\mathrm{2}} =\left(−\boldsymbol{{z}}\right)^{\mathrm{2}} \\ $$$$\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} −\boldsymbol{{z}}^{\mathrm{2}} =−\mathrm{2}\boldsymbol{{xy}} \\ $$$$\Rightarrow\left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} −\boldsymbol{{z}}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} \\ $$$$\Rightarrow\boldsymbol{{x}}^{\mathrm{4}} +\boldsymbol{{y}}^{\mathrm{4}} +\boldsymbol{{z}}^{\mathrm{4}} =\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{y}}^{\mathrm{2}} \boldsymbol{{z}}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}\boldsymbol{{z}}^{\mathrm{2}} \boldsymbol{{x}}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +\boldsymbol{{z}}^{\mathrm{4}} =\mathrm{2}\left(\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} \boldsymbol{{z}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} \boldsymbol{{x}}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{{x}}^{\mathrm{4}} +\boldsymbol{{y}}^{\mathrm{4}} +\boldsymbol{{z}}^{\mathrm{4}} \right)=\left(\boldsymbol{{xy}}\right)^{\mathrm{2}} +\left(\boldsymbol{{yz}}\right)^{\mathrm{2}} +\left(\boldsymbol{{zx}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}\boldsymbol{{xyz}}\left(\boldsymbol{{x}}+\boldsymbol{{y}}+\boldsymbol{{z}}\right) \\ $$$$\:\:\:\left[\because\:\boldsymbol{{x}}+\boldsymbol{{y}}+\boldsymbol{{z}}=\mathrm{0}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{{x}}^{\mathrm{4}} +\boldsymbol{{y}}^{\mathrm{4}} +\boldsymbol{{z}}^{\mathrm{4}} \right)=\left(\boldsymbol{{xy}}\right)^{\mathrm{2}} +\left(\boldsymbol{{yz}}\right)^{\mathrm{2}} +\left(\boldsymbol{{zx}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\boldsymbol{{xy}}.\boldsymbol{{yz}}+\mathrm{2}\boldsymbol{{xzyz}}+\mathrm{2}\boldsymbol{{xy}}.\boldsymbol{{zx}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{{x}}^{\mathrm{4}} +\boldsymbol{{y}}^{\mathrm{4}} +\boldsymbol{{z}}^{\mathrm{4}} \right)=\left(\boldsymbol{{xy}}+\boldsymbol{{yz}}+\boldsymbol{{zx}}\right)^{\mathrm{2}} \\ $$$$\therefore\frac{\mathrm{1}}{\mathrm{2}}\left\{\left(\boldsymbol{{b}}−\boldsymbol{{c}}\right)^{\mathrm{4}} +\left(\boldsymbol{{c}}−\boldsymbol{{a}}\right)^{\mathrm{4}} +\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)^{\mathrm{4}} \right\}= \\ $$$$\:\:\left\{\left(\boldsymbol{{b}}−\boldsymbol{{c}}\right)\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)+\left(\boldsymbol{{c}}−\boldsymbol{{a}}\right)\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)+\left(\boldsymbol{{b}}−\boldsymbol{{c}}\right)\left(\boldsymbol{{c}}−\boldsymbol{{a}}\right)\right\}^{\mathrm{2}} \\ $$$$\: \\ $$$$ \\ $$
Commented by Tawa11 last updated on 11/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$