begin{array}{c}sqrt{ } t^{3}+1 =left(frac{y^{4}+1}{y^{2}+1}right)^{5}end{array}

Question

begin{array}{c}sqrt{ } t^{3}+1  =left(frac{y^{4}+1}{y^{2}+1}right)^{5}end{array}

Accepted Answer

Step:1 ：Simplify the equation: (sqrt{t^3 + 1} = left(frac{y^4 + 1}{y^2 + 1}right)^5)Step:2 ：Solve for t in terms of y: (t = boxed{(-1 + frac{(y^4 + 1)^{10}}{(y^2 + 1)^{10}})^{frac{1}{3}}})

Answer

Step: 1 ：Simplify the equation: (sqrt{t^3 + 1} = left(frac{y^4 + 1}{y^2 + 1}right)^5)Step: 2 ：Solve for t in terms of y: (t = boxed{(-1 + frac{(y^4 + 1)^{10}}{(y^2 + 1)^{10}})^{frac{1}{3}}})

$\begin{array}{c}\sqrt{ } t^{3}+1 \\ =\left(\frac{y^{4}+1}{y^{2}+1}\right)^{5}\end{array}$

Answer

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