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

Explanation

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)`