Revision history [back]

I do not understand why

var('R','p','x')
sol1=solve(R^(1/3)*p == x^(1/3),x)
show(sol1)

this work but this fails to isolate $x$

var('R','p','x')
sol1=solve(R^(1/3)*p <= x^(1/3),x)
show(sol1)

And that this

var('R','p','x')
sol2=solve(0 == (R-x)^(1/3)*p + (1-p)* (-x)^(1/3),x)
show(sol2)

doesn't find $x$. And, as a consequence, of course that this :

var('R','p','x')
sol2=solve(0 <= (R-x)^(1/3)*p + (1-p)* (-x)^(1/3),x)
show(sol2)

fails. I have tried all simplification ways.

I do not understand why

var('R','p','x')
sol1=solve(R^(1/3)*p == x^(1/3),x)
show(sol1)

this work but this fails to isolate $x$

var('R','p','x')
sol1=solve(R^(1/3)*p <= x^(1/3),x)
show(sol1)

And that this

var('R','p','x')
sol2=solve(0 == (R-x)^(1/3)*p + (1-p)* (-x)^(1/3),x)
show(sol2)

doesn't find $x$. And, as a consequence, of course that cours this :

var('R','p','x')
sol2=solve(0 <= (R-x)^(1/3)*p + (1-p)* (-x)^(1/3),x)
show(sol2)

fails. I have tried all simplification ways.