Question

how many such number are their which are equal to their cube but not equal to their square​

Answer 1

Answer:

Yes, It is viable. Have a look :)

Let’s take a number ‘x'.

Now,

x=x^3

(x^3)-x=0

x(x^2-1)=0

On solving we get three values for x.

x=0 , 1, -1

Similarly, for

x=x^2

we have only two values for x.

x=0 & 1

Examining the above solution, it is luculent that x=-1 is the required number.

Apodictically it can be verified using basics of derivative and plotting a graph of these functions.

Step-by-step explanation:

please mark me brain list