Find all real numbers with the following property: the difference of its cube and its square is equal to the square of the difference of its square and the number itself.
Problem
Source: Juniors Problem 5
Tags: geometry, 3D geometry, quadratics, algebra, algebra unsolved
Bugi
30.07.2008 13:43
a^3-a^2=(a^2-a)^2
Then a^3-a^2=a^4+a^2-2a^3
3a^3=a^4+2a^2 (1).
Dividing by a^2 we get quadratic equation a^2-3a+2=0 which solutions are 2 and 1.
The other case is when we can't divide (1) by a^2 then a=0
mszew
30.07.2008 15:21
Be careful with a=0.
Bugi
30.07.2008 16:28
Thanks. I edited the upper post.
Taco12
21.07.2021 19:11
a3−a2=(a2−a)2, or a2(a2−3a+2)=0. Factoring further yields 0,1,2 as solutions.
OlympusHero
09.09.2021 00:22
Let the number be x. Then x3−x2=(x2−x)(x2−x). Note that x=0 works. Now dividing by x2 gives x−1=(x−1)2. Note that x=1 works. Now dividing by x−1 gives 1=x−1, so x=2 works. The answer is hence 0,1,2.