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Question Number 71073 by TawaTawa last updated on 11/Oct/19

Commented by Prithwish sen last updated on 11/Oct/19

$${\mathrm{a}}^2={\mathrm{b}}^2+{\mathrm{c}}^2+2\mathrm{bc}$$$$\therefore \mathbf{bc}=\frac{{\mathbf{a}}^2-{\mathbf{b}}^2-{\mathbf{c}}^2}{2}\Rightarrow {\mathbf{a}}^2-\mathbf{bc}=\frac{{\mathbf{a}}^2+{\mathbf{b}}^2+{\mathbf{c}}^2}{2}$$$$\mathbf{Similarly}{\mathbf{b}}^2-\mathbf{ca}=\frac{{\mathbf{a}}^2+{\mathbf{b}}^2+{\mathbf{c}}^2}{2}$$$${\mathbf{c}}^2-\mathbf{ab}=\frac{{\mathbf{a}}^2+{\mathbf{b}}^2+{\mathbf{c}}^2}{2}$$$$\therefore \mathbf{the}\mathbf{expression}$$$$\frac{2{\mathbf{a}}^2}{{\mathbf{a}}^2+{\mathbf{b}}^2+{\mathbf{c}}^2}+\frac{2{\mathbf{b}}^2}{{\mathbf{a}}^2+{\mathbf{b}}^2+{\mathbf{c}}^2}+\frac{2{\mathbf{c}}^2}{{\mathbf{a}}^2+{\mathbf{b}}^2+{\mathbf{c}}^2}=2$$

Commented by Henri Boucatchou last updated on 11/Oct/19

$$\mathit{N}\mathit{i}\mathit{c}\mathit{e}...$$

Commented by Prithwish sen last updated on 11/Oct/19

$$\mathbf{Thank}\mathbf{you}\mathbf{Sir}.$$

Commented by TawaTawa last updated on 11/Oct/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by $@ty@m123 last updated on 11/Oct/19

$$Followingmethodcanbeused$$$$tofindquicksolutionofMCQ:$$$$Leta=-1,b=0,c=1$$$$\therefore Givenexpression:$$$$1+0+1=2$$

Commented by MJS last updated on 11/Oct/19

$$\mathrm{but}\mathrm{then}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{valid}\mathrm{for}\mathrm{all}\mathrm{values}\mathrm{of}a,b,c$$$$\mathrm{with}a+b+c=0$$

Commented by $@ty@m123 last updated on 11/Oct/19

$$Canyougiveanexample?$$

Commented by $@ty@m123 last updated on 11/Oct/19

$$Ofcourse,$$$$thisisnotaproper\left(traditional\right)$$$$way,butinMCQtypetests,weneedtricky$$$$approachtosavetime.$$$$$$

Answered by MJS last updated on 11/Oct/19

$$c=-(a+b)$$$$\frac{a^2}{a^2+ab+b^2}+\frac{b^2}{a^2+ab+b^2}+\frac{{(a+b)}^2}{a^2+ab+b^2}=2$$$$$$

Commented by Prithwish sen last updated on 11/Oct/19

$$<\mathbf{Excellent}\mathbf{Sir}!$$

Commented by MJS last updated on 11/Oct/19

$$\mathrm{thank}\mathrm{you}$$

Commented by TawaTawa last updated on 11/Oct/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by peter frank last updated on 11/Oct/19

$$a+b+c=0$$$$a=-b-c$$$$a^2=-ab-ac....\left(i\right)$$$$a+b+c=0$$$$b=-a-c$$$$b^2=-ab-cb....\left(ii\right)$$$$a+b+c=0$$$$c=-a-b$$$$c^2=-ac-bc.....\left(iii\right)$$$$from$$$$\frac{a^2}{a^2-bc}+\frac{b^2}{b^2-ca}+\frac{c^2}{c^2-ab}....\left(iv\right)$$$$substitutesiiiiiiiniv$$$$a^2=-ab-ac....\left(i\right)$$$$b^2=-ab-cb....\left(ii\right)$$$$c^2=-ac-bc.....\left(iii\right)$$$$\frac{ab+ca}{ab+ca+bc}+\frac{ba+bc}{ab+ca+bc}+\frac{ca+cb}{ab+ca+bc}$$$$\frac{2ab+2ac+2bc}{ab+ca+bc}$$$$2$$$$$$

Commented by TawaTawa last updated on 11/Oct/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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