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Question Number 12623 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/Apr/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/Apr/17

one or more is true.find it or them!

$${one}\:{or}\:{more}\:{is}\:{true}.{find}\:{it}\:{or}\:{them}! \\ $$

Answered by prakash jain last updated on 27/Apr/17

(i) (a−b)^2 +(ab+1)^2 =a^2 +b^2 +a^2 b^2 +1 =a^2 (b^2 +1)+(b^2 +1)=(a^2 +1)(b^2 +1) (ii) (a+b)^2 +(ab−1)^2 =a^2 +b^2 +a^2 b^2 +1 =(a^2 +1)(b^2 +1) (iii) (a+b)^2 +(ab+1)^2 =a^2 +b^2 +a^2 b^2 +1+4ab ≠(a^2 +1)(b^2 +1)

$$\left({i}\right) \\ $$$$\left({a}−{b}\right)^{\mathrm{2}} +\left({ab}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{1} \\ $$$$={a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +\mathrm{1}\right)+\left({b}^{\mathrm{2}} +\mathrm{1}\right)=\left({a}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\left({ii}\right) \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} +\left({ab}−\mathrm{1}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{1} \\ $$$$=\left({a}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\left({iii}\right) \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} +\left({ab}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{1}+\mathrm{4}{ab} \\ $$$$\neq\left({a}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}} +\mathrm{1}\right) \\ $$

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