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[Post New]10 Mar 2008 19:03:08 IST
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hsbhatt (4460)

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Olaaa!! Perrrfect answer. 840  [970 rates]
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\text{No calculus-valculus required}\\ \\ \text {The given expression is equivalent to} \\ \\ (x^8-2x^4+1) + (x^6-2x^3+1)+(x^4-2x^2+1)+(x^2-2x+1)+5 \\ \\ = (x^4-1)^2 + (x^3-1)^2 + (x^2-1)^2+(x-1)^2+5 \\ \\ \text {Now, it is easy to see that the minimum value is 5, attained at x=1}
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