Question

Solve for the following equation . x^6-9x^3+10=0

Answer 1

• x = ∛10
• x = -∛10/2 +(∛10√3)/2i
• x = -∛10/2 -(∛10√3)/2i
• x = -1
• x = 1/2 -(√3)/2i
• x = 1/2 +(√3)/2i

Explanation:

You want to solve the 6th-degree equation x^6 -9x^3 +10 = 0.

Quadratic

The equation is basically quadratic in x^3. Substituting z = x^3, we have ...

z^2 -9z +10 = 0

(z -10)(z +1) = 0 . . . . . factored form

z = 10, z = -1 . . . . . . . values of z that make the factors zero

Cubics

Now, we have two cubic equations to solve:

• x^3 = 10 ⇒ x = ∛10
• x^3 = -1 ⇒ x = ∛(-1) = -1

The real root is given by the cube root function. The imaginary roots are that value multiplied by 1∠±120° = (-1/2 ±(√3)/2i). Then the six roots are ...

• x = ∛10
• x = -∛10/2 +(∛10√3)/2i
• x = -∛10/2 -(∛10√3)/2i
• x = -1
• x = 1/2 -(√3)/2i
• x = 1/2 +(√3)/2i

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Additional comment

The product ∛10√3 can be simplified to ...

`\sqrt[3]{10}√(3)=(10^(2/6))(3^(3/6))=(10^23^3)^(1/6)=\sqrt[6]{2700}\approx 3.7315903`