Factorize `(i) (x^4+4)` `(ii) (x^2+(4)/(x^2))` `(iii) (x^4+(1)/(x^4)+1

1.	Factorize `(i) (x^4+4)` `(ii) (x^2+(4)/(x^2))` `(iii) (x^4+(1)/(x^4)+1)`
Answer» We have `(i) (x^4+4)` `=(x^2)^2+2^2+(2xx x^2xx 2)-(2xx x^2xx2) " "["adding and subtracting" (2xx x^2xx2)]` `=(x^2+2)^2-4x^2` `=(x^2+2)^2-(2x)^2` `(x^2+2-2x)(x^2+2x+2x) =(x^2-2x+2)=(x^2-2x+2)(x^2+2x+2)` `therefore (x^4+4)=(x^2-2x+2)(x^2+2x+2)`. ` (ii) (x^2+(4)/(x^2))` `=x^2+((2)/(x))^2+(2xx x xx (2)/(x))-(2 xx x xx (2)/(x))` `=(x+(2)/(x))^2-4={(x+(2)/(x))^2-2^2}` `=(x+(2)/(x)-2)(x+(2)/(x)+2)`. ` therefore (x^2+(4)/(x^2))=(x+(2)/(x)-2)(x+(2)/(x)+2)`. `(iii) (x^4+(1)/(x^4)+1)` `=(x^4+(1)/(x^4)+2)-1=(x^2+(1)/(x^2))^2-1^2` `=(x^2+(1)/(x^2)-1){(x^2+(1)/(x^2)+2)-1}` `=(x^2+(1)/(x^2)-1){(x+(1)/(x))^2-1^2}` `=(x^2+(1)/(x^2)-1)(x+(1)/(x)-1)(x+(1)/(x)+1)` ` therefore (x^4+(1)/(x^4)+1)=(x^2+(1)/(x^2)-1)(x+(1)/(x)-1)(x+(1)/(x)=1)`

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