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Question Number 110268 by Khanacademy last updated on 28/Aug/20

Commented by som(math1967) last updated on 28/Aug/20

$$-2$$

Commented by Khanacademy last updated on 28/Aug/20

$$?$$

Commented by som(math1967) last updated on 29/Aug/20

$$(\frac{{\mathrm{a}}^2}{\mathrm{b}+\mathrm{c}}+\mathrm{a})+(\frac{{\mathrm{b}}^2}{\mathrm{c}+\mathrm{a}}+\mathrm{b})+(\frac{{\mathrm{c}}^2}{\mathrm{c}+\mathrm{a}}+\mathrm{c})-(\mathrm{a}+\mathrm{b}+\mathrm{c})$$$$\frac{\mathrm{a}(\mathrm{a}+\mathrm{b}+\mathrm{c})}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}(\mathrm{a}+\mathrm{b}+\mathrm{c})}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}(\mathrm{a}+\mathrm{b}+\mathrm{c})}{\mathrm{a}+\mathrm{b}}$$$$-(\mathrm{a}+\mathrm{b}+\mathrm{c})$$$$(\mathrm{a}+\mathrm{b}+\mathrm{c})(\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}})$$$$-(\mathrm{a}+\mathrm{b}+\mathrm{c})$$$$(\mathrm{a}+\mathrm{b}+\mathrm{c})(-1)-(\mathrm{a}+\mathrm{b}+\mathrm{c})$$$$[\because \frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}=-1]$$$$-2(\mathrm{a}+\mathrm{b}+\mathrm{c})$$$$\therefore (\frac{{\mathrm{a}}^2}{\mathrm{b}+\mathrm{c}}+\frac{{\mathrm{b}}^2}{\mathrm{c}+\mathrm{a}}+\frac{{\mathrm{c}}^2}{\mathrm{a}+\mathrm{b}}):(\mathrm{a}+\mathrm{b}+\mathrm{c})$$$$-2(\mathrm{a}+\mathrm{b}+\mathrm{c}):(\mathrm{a}+\mathrm{b}+\mathrm{c})$$$$-2[\mathrm{If}(\mathrm{a}+\mathrm{b}+\mathrm{c})\ne 0]$$

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