Solve `(x^3-4x)sqrt(x^2-1)=0.`

Explanation

Given equation is solvable for `x^(2)- 1 ge 0`
or x `in(-infty,-1]cup[1, infty)`
Now, `(x^(3) - 4x)sqrt(x^(2)-1 )=0`.
or `x(x-2)(x+2)sqrt(x^(2)-1) = 0`
or `x = 0, -2, 2, -1, 1`
But x `in(-infty, - 1]cup[1,infty)`
`rArr" "x = pm 1, pm 2`
Clearly for x = 0 , `sqrt(x^(2)-1)` is not defined