Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 173993 by AgniMath last updated on 22/Jul/22

Answered by som(math1967) last updated on 22/Jul/22

$$xy+yz+zx+2xyz$$$$=\frac{ab}{(b+c)(c+a)}+\frac{bc}{(c+a)(a+b)}$$$$+\frac{ca}{(a+b)(b+c)}+\frac{2abc}{(a+b)(b+c)(c+a)}$$$$=\frac{ab(a+b)+bc(b+c)+ca(c+a)+2abc}{(a+b)(b+c)(c+a)}$$$$=\frac{(a+b)(b+c)(c+a)}{(a+b)(b+c)(c+a)}★$$$$=1$$$$★ab(a+b)+b^{2}c+{bc}^2+c^{2}a+{ca}^2+2abc$$$$=ab(a+b)+c(a^2+b^2+2ab)+c^2(a+b)$$$$=ab(a+b)+c{(a+b)}^2+c^2(a+b)$$$$=(a+b)(ab+ca+bc+c^2)$$$$=(a+b)(b+c)(c+a)$$

Commented by Tawa11 last updated on 23/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com