Question 544648
okay, took me a few tims thru but got the answer. Thanks to JBarnum for getting me thinking of factoring.

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{{{sqrt( 1 + (x^3 - 1/(4x^3))^2)}}}
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factor (a - b)^2 = a^2 - 2ab - b^2
{{{sqrt( 1 + (x^3)^2 - 2x^3(1/(4x^3)) + ( 1/(4x^3))^2)}}}
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{{{sqrt(1 + x^6 - 1/2 + 1/(16x^6))}}}

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combine like terms
{{{sqrt(x^6 + 1/(16x^2) + 1/2)}}}
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multiply thru by x^6/x^6 to get rid of the x^6 in the denominator
{{{sqrt((x^12 + x^6/2 + 1/16)(1/x^6))}}}
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let U = x^6

{{{sqrt((U^2 + U/2 + 1/16)(1/U))}}}
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Factor
{{{sqrt((U + 1/4)^2 (1/U))}}}
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{{{(U + 1/4) * sqrt(1/U)}}}
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bring back in the X^6 in place of U.
{{{(x^6 + 1/4)* sqrt(1/x^6)}}}
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{{{(x^6+ 1/4)/x^3}}}
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Multiply thru by 4/4 to get rid of 1/4 fraction
{{{(4x^6 + 1)/ 4x^3}}}
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Simplify
{{{(4x^6)/(4x^3) + 1/(4x^3)}}}
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{{{x^3 = 1/(4x^3)}}} Solution