Solve the equation cube root (x^3 - 3x^2) + 1 = x.  

Expert Answers
hala718 eNotes educator| Certified Educator

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

First we will subtract 1 from both sides.

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

Now we will cube both sides.

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

Now we will open the brackets.

==> x^3 - 3x^2  = x^3 -3x^2 + 3x -1

Now we will combine like terms.

==> x^3 - x^2 - 3x^2 + 3x^2 - 3x + 1 = 0

==> 0 - 0 - 3x + 1 = 0

==> -3x = -1

Now we will divide by -3.

==> x = -1/-3 = 1/3

==> x= 1/3

justaguide eNotes educator| Certified Educator

We have the equation (x^3 - 3x^2) ^ (1/3) + 1 = x

subtract 1 from both the sides

=> (x^3 - 3x^2) ^ (1/3) = x – 1

cube both the sides

=> (x^3 - 3x^2) ^ (1/3) ^3 = (x – 1) ^3

=> x^3 - 3x^2 = x^3 – 3x^2 + 3x -1

=> x^3 - 3x^2 - x^3 + 3x^2 - 3x + 1 = 0

cancel the common terms

=> - 3x + 1 = 0

=> 3x = 1

=> x = 1/3

Therefore x is equal to 1/3.

neela | Student

To solve the equation cube root (x^3 - 3x^2) + 1 = x.

=> {x-3x}^(1/3) = (x-1).

= > (x^2-3x)^(1/3) = (x-1).

We raise to the 3rd power both sides:

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

x^3-3x^2 = x^3-3x^2+3x-1- x^3+3x^2

0 = 3x-1.

So x = 1/3.