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i think we need to consider the three cases in parallel eg: if X<0 what will happen to other two expressions
If (x^3) is negative, x < 0 say X is -0.5, X^3 is -ive , 1-X is +ive and 1+X too is positive. expression holds. Now say X=-2, X^3 is -ive , 1-X is +ive but 1+X is -ive . expression doesnt hold . so we cannot say that if X<0 , X^3 (1-X) (1+X) < 0
1+X will be -ive only when X <-1 , in that case 1-X will be +ive and X^3 will be -ive , over all expression will be +ive so this set of values of X also ruled out.
now 1-X will be -ive for all values of X greater than 1 and the other two X^3 and 1+X will be positive in this case. we can safely say that for all X>1 the expression will hold.
may be i am wrong though in my thought process , pls advise.
(1) x^3(1-x^2)<0 --> x^3(1-x)(1+x)<0 --> the roots are -1, 0, and 1 (equate the expressions to zero to get the roots and list them in ascending order), this gives us 4 ranges: x<-1, -1<x<0, 0<x<1, and x>1.
Now, test some extreme value: for example if x is very large number then x and 1+x are positive, while 1-x is negative, which gives negative product for the whole expression, so when x>1 the expression is negative. Now the trick: as in the 4th range expression is negative then in 3rd it'll be positive, in 2nd it'll be negative again and finally in 1st it'll be positive: + - + -. So, the ranges when the expression is negative are: -1<x<0 and x>1.
So, x could be negative as well as positive. Not sufficient.
Or: x^3(1-x^2)<0 --> x(1-x^2)<0 --> x<x^3 --> -1<x<0 or x>1. Not sufficient.
(2) x^2-1<0 --> x^2<1 --> -1<x<1. Not sufficient.
(1)+(2) Intersection of the ranges from (1) and (2) is -1<x<0, hence the answer to the question whether x<0 is YES. Sufficient.