What is ((2t^2+1)/(t^3-t)-x)*(1/(t-t^3)-x) ?

The solution to ((2t^2+1)/(t^3-t)-x)*(1/(t-t^3)-x) is -((-t^3x+tx+2t^2+1)(-xt+xt^3+1))/((t^3-t)^2)

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simplify ((2t^2+1)/(t^3-t)-x)*(1/(t-t^3)-x)

simplify

(\frac{2 t ^{2} + 1}{t ^{3} - t} - x) \cdot (\frac{1}{t - t ^{3}} - x)

- \frac{( - t ^{3} x + t x + 2 t ^{2} + 1 ) ( - x t + x t ^{3} + 1 )}{( t ^{3} - t ) ^{2}}

Solution steps

(\frac{2 t ^{2} + 1}{t ^{3} - t} - x) (\frac{1}{t - t ^{3}} - x)

Join

\frac{2 t ^{2} + 1}{t ^{3} - t} - x : \frac{- t ^{3} x + t x + 2 t ^{2} + 1}{t ^{3} - t}

= \frac{- t ^{3} x + t x + 2 t ^{2} + 1}{t ^{3} - t} (\frac{1}{t - t ^{3}} - x)

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( - t ^{3} x + t x + 2 t ^{2} + 1 ) ( \frac{1}{t - t ^{3}} - x )}{t ^{3} - t}

Simplify

(- t^{3} x + t x + 2 t^{2} + 1) (\frac{1}{t - t ^{3}} - x) : \frac{( - t ^{3} x + t x + 2 t ^{2} + 1 ) ( - x t + x t ^{3} + 1 )}{t - t ^{3}}

= \frac{\frac{( - t ^{3} x + t x + 2 t ^{2} + 1 ) ( - x t + x t ^{3} + 1 )}{t - t ^{3}}}{t ^{3} - t}

Apply the fraction rule:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

= \frac{( - t ^{3} x + t x + 2 t ^{2} + 1 ) ( - x t + x t ^{3} + 1 )}{( t - t ^{3} ) ( t ^{3} - t )}

Simplify

(t - t^{3}) (t^{3} - t) : - (t^{3} - t)^{2}

= \frac{( - t ^{3} x + t x + 2 t ^{2} + 1 ) ( - x t + x t ^{3} + 1 )}{- ( t ^{3} - t ) ^{2}}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{( - t ^{3} x + t x + 2 t ^{2} + 1 ) ( - x t + x t ^{3} + 1 )}{( t ^{3} - t ) ^{2}}

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What is ((2t^2+1)/(t^3-t)-x)*(1/(t-t^3)-x) ?
The solution to ((2t^2+1)/(t^3-t)-x)*(1/(t-t^3)-x) is -((-t^3x+tx+2t^2+1)(-xt+xt^3+1))/((t^3-t)^2)