Final answer to the problem
$\frac{\sqrt[3]{x^6+8}\left(4x^2-\sqrt{3x^4+1}\right)}{\left(4x^2\right)^2-3x^4-1}$
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Step-by-step Solution
Learn how to solve factor by difference of squares problems step by step online. Rationalize and simplify the expression ((x^6+8)^(1/3))/(4x^2+(3x^4+1)^(1/2)). Multiply and divide the fraction \frac{\sqrt[3]{x^6+8}}{4x^2+\sqrt{3x^4+1}} by the conjugate of it's denominator 4x^2+\sqrt{3x^4+1}. Multiplying fractions \frac{\sqrt[3]{x^6+8}}{4x^2+\sqrt{3x^4+1}} \times \frac{4x^2-\sqrt{3x^4+1}}{4x^2-\sqrt{3x^4+1}}. Solve the product of difference of squares \left(4x^2+\sqrt{3x^4+1}\right)\left(4x^2-\sqrt{3x^4+1}\right).
Final answer to the problem
$\frac{\sqrt[3]{x^6+8}\left(4x^2-\sqrt{3x^4+1}\right)}{\left(4x^2\right)^2-3x^4-1}$
