ksaimouli
y^4+(1/16y^4)+1/2



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ksaimouli
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\[y^4+\frac{ 1 }{ 16y^4 }+\frac{ 1 }{ 2 }\]

ksaimouli
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perfect square

ksaimouli
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@Jonask

phi
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you could make 16y^4 the common denominator and add up the terms
the denominator is a perfect square. Does the numerator factor?

ksaimouli
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is their any easy way instead of just taking common denominator

ksaimouli
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@zepdrix

ksaimouli
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i know a^2+2ab+b^2

phi
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It is not that hard

phi
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16 y^8 + 1 + 8 y^4

ksaimouli
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=(a+b)^2

phi
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rearrange as
16 y^8 + 8 y^4 + 1
if you let y^4 = x, this is also
16x^2 + 8x +1

Jonask
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i just tried\[(y^2+\frac{1}{4y^2})^2\]

phi
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if it is square your only hope is (4x+1)^2 which works

ksaimouli
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is that =0 because what happened to common denominator

ksaimouli
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(4x^4+1)^2

Jonask
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is it not true to say\[(y^2+\frac{1}{4y^2})^2=y^4+2(y^2)(\frac{1}{4y^2})+\frac{1}{16y^4}=y^4+\frac{1}{2}+\frac{1}{16y^4}\]