Solving a third degree equation
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Solve the equation
x^3 - 3x^2 + 3 = 0
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Solution Summary
I show how to solve general third degree equations. I work out the details explicitely for the equation x^3 - 3x^2 + 3 = 0
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To solve an equation of the form:
x^3 + a x^2 + b x + c = 0
follow the following steps.
First, get rid of the quadratic term a x^2 using the substitution:
x = y - a/3
This step is analogous to how you solve the quadratic equation when you write it as a perfect square.
In this case we aren't finshed yet as the equation now is of the form:
y^3 + p y + q = 0
How do we solve this equation?
Consider the identity:
(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
We can rewrite the right hand side as follows:
a^3 + 3a^2b + 3ab^2 + b^3 =
a^3 + b^3 + 3ab(a+b)
So, we have:
(a+b)^3 = 3ab(a+b)+ a^3 + b^3 ---->
(a+b)^3 - 3ab(a+b) -([a^3+b^3) = 0
Of course, this equation is always ...
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