Solve a linear equation

1. Dec 4, 2015

science_rules

(x - 2)^3 = x^2(x - 6)
= (x^3 - 8) = (x^3 - 6x^2)

(x^3 - 8) = (x^3 - 6x^2)
-x^3 -x^3

= -8 = -6x^2

I am not sure how to solve this for x

2. Dec 4, 2015

Samy_A

A clue that something went wrong is that $-8=-6x²$ is not a linear equation.
Note that, in general, $(a-b)³ \neq a^³-b^³$, as you seem to think in the expression I set in red.

Last edited: Dec 4, 2015
3. Dec 4, 2015

science_rules

(x - 2)^3 = x^2(x - 6)
= (x-2)(x-2)(x-2) = x^3 - 6x^2
= (x^2 - 2x -2x + 4) (x-2)
= (x^2 - 4x + 4)(x-2)
= x^3 - 6x^2 + 12x -8
x^3 - 6x^2 + 12x -8 = x^3 -6x^2 ???

4. Dec 4, 2015

science_rules

I know I am supposed to even out the inequality by subtracting and adding to each side but I am still not sure how to solve for x

5. Dec 4, 2015

Samy_A

You can start by removing the terms that appear on both sides of the equation. Then see what remains.

6. Dec 4, 2015

science_rules

x^3 - 6x^2 + 12x -8 = x^3 -6x^2
= x^3 + 12x -8 = x^3
subtract 12x from both sides
= x^3 -8 = x^3 - 12x
subtract x^3 from both sides
= -8 = -12x
x = -8/12
And I believe I got the answer

7. Dec 4, 2015

science_rules

I meant: x = -8/-12

8. Dec 4, 2015

Samy_A

Correct. You can simplify this to 8/12 or 2/3.

9. Dec 4, 2015

Staff: Mentor

Don't connect pairs of equations with '='. An equation is not "equal to" another equation. One equation can be equivalent to another equation (same solution set for both) or one equation can imply another equation. There are separate symbols for each of these relationships, but in your case it would be OK to not use any symbol at all between the successive equations.
You should never leave an answer like this. -8/-12 is the same as 8/12, which can be simplified to 2/3.

Also, this question has nothing to do with linear algebra or abstract algebra, so I moved it to the general math section.

10. Dec 6, 2015

HallsofIvy

Staff Emeritus
In addition to everything else this is NOT a linear equation. This might have been in a section on "linear equations" because after canceling the "$x^3- 6x^2$" terms, the remaining equation is linear.