We have `x^3y + xy-y = 0`

A trivial solution is `y=0`, `x in RR`

Otherwise we need to solve

`x^3 + x - 1 =0`

Draw the graph

The only real (`x in RR`) solution to this is positive,

therefore we can let `x =...

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We have `x^3y + xy-y = 0`

A trivial solution is `y=0`, `x in RR`

Otherwise we need to solve

`x^3 + x - 1 =0`

Draw the graph

The only real (`x in RR`) solution to this is positive,

therefore we can let `x = e^t` , which is always positive.

Use a Taylor series approximation to the function `f(t) = e^(3t) + e^t -1`

Now, using standard Taylor series approximations

`e^(3t) approx 1 + 3t + (3t)^2/2` and

`e^t approx 1 + t + t^2/2`

Therefore `e^(3t) + e^t -1 approx 2 + 4t + (10t^2)/2 -1 = 1 +4t + 5t^2`

We now want to solve

`5t^2 + 4t + 1 approx 0`

Using the quadratic formula

`implies` `t approx (-4 +- sqrt(16 - 20))/10 = -2/5 + r`

If the solution is real (it must be to insure `x` is real) and unique then `r approx 0`

and we have that `t approx -2/5`

`implies` `x approx e^(-2/5) = 0.67`

Check

`x^3 + x -1 approx 0.67^3 +0.67 -1 = -0.028`

which is approximately zero.

**Therefore ` `the possible solutions are**

`y=0`, `x in RR` or `x approx 0.67`,`y in RR`