x^4-x<0
How do I solve it?
Maybe you can factor $\displaystyle \displaystyle x^3-1$ further?
Hint: $\displaystyle \displaystyle a^3-b^3 = (a-b)(a^2-ab+b^2)$
Sketching the graph of $\displaystyle f(x) = x^4-x$ will help after you have found the zeros.
$\displaystyle x^4 - x < 0$
$\displaystyle x(x^3 - 1) < 0$
$\displaystyle x(x-1)(x^2 + x + 1) < 0$
note that the factor $\displaystyle (x^2 + x + 1) > 0$ for all x (how can you check that the factor is always positive?)
for the other two factors that can equal zero ... critical values are x = 0 and x = 1
these two critical values divide the x-value number line into three regions ...
x < 0 , 0 < x < 1 , and x > 0
test a single x-value from each of the above intervals into the original inequality to see if this value makes the inequality true or false ...
if true, then all values of x in that interval make the inequality true ... if false, then otherwise.