0

If

EDIT 1:

typo edited

$$ 2 (x^6 + y^6 ) - 3(x^4 + y^4 ) + (x^2+y^2) = 0,$$

then show that

$$ (x^2+y^2) =1 $$

The trigonometric converse is known already.

Narasimham
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  • This is the converse of http://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion – lab bhattacharjee Mar 23 '16 at 15:16
  • Seeing as this question seemingly stemmed from http://math.stackexchange.com/questions/1710182/simplify-2-sin6x-cos6x-3-sin4-x-cos4-x-1 , maybe the statement isn't true beyond the range of the trig functions? – Airdish Mar 23 '16 at 15:28
  • @lab bhattacharjee I fail to see any connections, please say how so. – Narasimham Mar 23 '16 at 15:43
  • If true fo trig it is true for polynomial also, right? no matter its source/origin... – Narasimham Mar 24 '16 at 03:15
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1 Answers1

1

False. $2(x^6+y^6)-3(x^4+y^4)+(x^2+y^2)=(x^2+y^2-1)(2x^4+2y^4-2x^2y^2-x^2-y^2)$ which explains why the reverse is true. Take $x=y=0$ then given relation is true and $x^2+y^2=0$.

Ng Chung Tak
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almagest
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