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ksaimouliBest ResponseYou've already chosen the best response.0
\[y^4+\frac{ 1 }{ 16y^4 }+\frac{ 1 }{ 2 }\]
 one year ago

phiBest ResponseYou've already chosen the best response.0
you could make 16y^4 the common denominator and add up the terms the denominator is a perfect square. Does the numerator factor?
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.0
is their any easy way instead of just taking common denominator
 one year ago

phiBest ResponseYou've already chosen the best response.0
rearrange as 16 y^8 + 8 y^4 + 1 if you let y^4 = x, this is also 16x^2 + 8x +1
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
i just tried\[(y^2+\frac{1}{4y^2})^2\]
 one year ago

phiBest ResponseYou've already chosen the best response.0
if it is square your only hope is (4x+1)^2 which works
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.0
is that =0 because what happened to common denominator
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
is it not true to say\[(y^2+\frac{1}{4y^2})^2=y^4+2(y^2)(\frac{1}{4y^2})+\frac{1}{16y^4}=y^4+\frac{1}{2}+\frac{1}{16y^4}\]
 one year ago
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