# Solve the equation x^6-4x^3+3=0

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We have to solve x^6 - 4x^3 + 3 = 0

Let y = x^3

x^6 - 4x^3 + 3 = 0

=> y^2 - 4y + 3 = 0

=> y^2 - 3y - y + 3 = 0

=> y(y - 3) - 1(y - 3) = 0

=> (y - 1)(y-3) = 0

We get y = 1 and y = 3

Now y = x^3

Therefore we have

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

x^3 = 3 => x = 3^(1/3) = cube root 3

The values are:

**x = 1 **

**x = cube root 3**

The equation could be solvd using substitution technique. We'll reduce the equatin to a quadratic.

We'll note x^3 = t

t^2 - 4t + 3 = 0

We'll apply the quadratic formula:

t1 = [4+sqrt(16 - 12)]/2

t1 = (4+2)/2

t1 = 3

t2 = (4-2)/2

t2 = 1

But x^3 = t

x^3 = 3

x = 3^(1/3)

x^3 = 1

x = 1

**The solutions of the equation are: {3^(1/3); 1 }.**