how do you do this math equation step by step: (x-1)^3 - (x-1)^2 =0?

Answer 1

taking (x-1)^2 common
[(x-1)^2](x-1-1)=0
(x-1)^2)(x-2)=0
so x=1,1 or 2

Answer 2

Everyone else is doing it the right, mathematical way.

I'd like to point out one other way, though.

Call z = x - 1, for a moment. Then you have:

z³ - z² = 0
z³ = z²

There are only two numbers where you get the same thing when you square them as when you cube them, 1³ = 1² = 1 and 0³ = 0² = 0.

So if z = 1, then 1 = x - 1, so x = 2.

And if z = 0, then 0 = x - 1, so x = 1.

Just another way of looking at it.

Answer 3

First notice that both terms have at least 2 factors of (x-1) in common. This means that we can take them out as follows:

(x-1)^3 - (x-1)^2 = [(x-1)^2][(x-1) - 1]

Simplifying the RHS and setting it equal to 0, we get:

[(x-1)^2][x-2] = 0

So we have that x = 1 OR x = 2

Answer 4

First factor out an (x - 1) term:
(x - 1)[ (x - 1)² - (x - 1) ] = 0

And then again:
(x - 1)(x - 1) [ (x - 1) - 1 ] = 0

Now simplify the stuff in the brackets:
(x - 1)(x - 1)(x - 2) = 0

So the answers are x = 1 or x = 2

Answer 5

Since when you subtract they equal 0, you could always make them equal to each other.

(x-1)^3 = (x-1)^2
(x-1)(x-1)(x-1) = (x-1)(x-1)
x^3 + 1 = x^2 - 2x +1
x^3 = x^2 -2x
Factor out an x:
x^2(x^2 = x - 2)
x^2 = x - 2
x = (x-2)^1/2

Answer 6

ok first u foil everything out

solve for the latter part of the equation primarily.
u get x^2-2x+1. then u multiply one (x-1) term to that and you get
x^3-3x^2+x-1subtract the two equations and solve for x

Answer 7

(x-1)^3 - (x-1)^2 =0
(x-1)^2[x-1-1] = 0
(x-1)^2(x-2)= 0
x-2 = 0
x=2
(x-1)^2=0
x-1=0
x=1
x=1 is a double root

Answer 8

(x-1)^3 = (x-1)^2

[(x-1)^3] / [(x-1)^2] = [(x-1)^2] / [(x-1)^2]

(x-1) = 1

x= 1+1

x=2

Answer 9

1st (x-1)(x-1)(x-1)-(x-1)(x-1)=0
2nd you do foil... x-1(x-1)=x^2- 2x+1
3rd u rewrite it... (X-1)(x^2-2x+1)-(X^2-2x+1)=0
4th

Answer 10

i dont feel like typing it all out