3 x 3 x 3 x 3 x3

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "x3"   was replaced by   "x^3".  1 more similar replacement(s).

Step by step solution :

Step  1  :

Equation at the end of step  1  :

Step  2  :

Checking for a perfect cube :

 2.1    x4-3x3-x+3  is not a perfect cube

Trying to factor by pulling out :

 2.2      Factoring:  x4-3x3-x+3

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  -x+3 
Group 2:  x4-3x3

Pull out from each group separately :

Group 1:   (-x+3) • (1) = (x-3) • (-1)
Group 2:   (x-3) • (x3)
               -------------------
Add up the two groups :
               (x-3)  •  (x3-1) 
Which is the desired factorization

Trying to factor as a Difference of Cubes:

 2.3      Factoring:  x3-1

Theory : A difference of two perfect cubes,  a3 - b3 can be factored into
              (a-b) • (a2 +ab +b2)

Proof :  (a-b)•(a2+ab+b2) =
            a3+a2b+ab2-ba2-b2a-b3 =
            a3+(a2b-ba2)+(ab2-b2a)-b3 =
            a3+0+0-b3 =
            a3-b3

Check :  1  is the cube of   1 
Check :  x3 is the cube of   x1

Factorization is :
             (x - 1)  •  (x2 + x + 1)

Trying to factor by splitting the middle term

 2.4     Factoring  x2 + x + 1

The first term is,  x2  its coefficient is  1 .
The middle term is,  +x  its coefficient is  1 .
The last term, "the constant", is  +1 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 1 = 1

Step-2 : Find two factors of  1  whose sum equals the coefficient of the middle term, which is   1 .

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Equation at the end of step  2  :

Step  3  :

Theory - Roots of a product :

 3.1    A product of several terms equals zero. When a product of two or more terms equals zero, then at least one of the terms must be zero. We shall now solve each term = 0 separately In other words, we are going to solve as many equations as there are terms in the product Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

 3.2      Solve  :    x-1 = 0 Add  1  to both sides of the equation : 
                      x = 1

Parabola, Finding the Vertex :

 3.3      Find the Vertex of   y = x2+x+1Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 1 , is positive (greater than zero). Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions. Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex. For any parabola,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is  -0.5000  Plugging into the parabola formula  -0.5000  for  x  we can calculate the  y -coordinate : 
  y = 1.0 * -0.50 * -0.50 + 1.0 * -0.50 + 1.0
or   y = 0.750

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x2+x+1
Axis of Symmetry (dashed)  {x}={-0.50} 
Vertex at  {x,y} = {-0.50, 0.75} 
Function has no real roots

Solve Quadratic Equation by Completing The Square

 3.4     Solving   x2+x+1 = 0 by Completing The Square .Subtract  1  from both side of the equation :
   x2+x = -1

Now the clever bit: Take the coefficient of  x , which is  1 , divide by two, giving  1/2 , and finally square it giving  1/4

Add  1/4  to both sides of the equation :
  On the right hand side we have :
   -1  +  1/4    or,  (-1/1)+(1/4) 
  The common denominator of the two fractions is  4   Adding  (-4/4)+(1/4)  gives  -3/4 
  So adding to both sides we finally get :
   x2+x+(1/4) = -3/4

Adding  1/4  has completed the left hand side into a perfect square :
   x2+x+(1/4)  =
   (x+(1/2)) • (x+(1/2))  =
  (x+(1/2))2
Things which are equal to the same thing are also equal to one another. Since
   x2+x+(1/4) = -3/4 and
   x2+x+(1/4) = (x+(1/2))2
then, according to the law of transitivity,
   (x+(1/2))2 = -3/4

We'll refer to this Equation as  Eq. #3.4.1  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
   (x+(1/2))2  is
   (x+(1/2))2/2 =
  (x+(1/2))1 =
   x+(1/2)

Now, applying the Square Root Principle to  Eq. #3.4.1  we get:
   x+(1/2) = √ -3/4

Subtract  1/2  from both sides to obtain:
   x = -1/2 + √ -3/4
In Math,  i  is called the imaginary unit. It satisfies   i2  =-1. Both   i   and   -i   are the square roots of   -1 

Since a square root has two values, one positive and the other negative
   x2 + x + 1 = 0
   has two solutions:
  x = -1/2 + √ 3/4 •  i 
   or
  x = -1/2 - √ 3/4 •  i 

Note that  √ 3/4 can be written as
  √ 3  / √ 4   which is √ 3  / 2

Solve Quadratic Equation using the Quadratic Formula

 3.5     Solving    x2+x+1 = 0 by the Quadratic Formula .According to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B2-4AC
  x =   ————————
                      2A   In our case,  A   =     1
                      B   =    1
                      C   =   1 Accordingly,  B2  -  4AC   =
                     1 - 4 =
                     -3Applying the quadratic formula :

               -1 ± √ -3
   x  =    —————
                    2In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written  (a+b*i) 

Both   i   and   -i   are the square roots of minus 1

Accordingly,√ -3  = 
                    √ 3 • (-1)  =
                    √ 3  • √ -1   =
                    ±  √ 3  • i

  √ 3   , rounded to 4 decimal digits, is   1.7321
 So now we are looking at:
           x  =  ( -1 ±  1.732 i ) / 2

Two imaginary solutions :

Solving a Single Variable Equation :

 3.6      Solve  :    x-3 = 0 Add  3  to both sides of the equation : 
                      x = 3

Four solutions were found :

1.  x = 3
2.  x =(-1-√-3)/2=(-1-i√ 3 )/2= -0.5000-0.8660i
3.  x =(-1+√-3)/2=(-1+i√ 3 )/2= -0.5000+0.8660i
4.  x = 1