# Find H

#### TheJason

Staff member
Moderator
There should be three answers (roots) (post edited).

$$\displaystyle 15 + 3\,h^{3} = 16 - 9\,h^{3}$$

$$\displaystyle 15 -15 + 3\,h^{3} = 16 - 15 - 9\,h^{3}$$

$$\displaystyle 3\,h^{3} = 1 - 9\,h^{3}$$

$$\displaystyle 3\,h^{3} + 9\,h^{3} = 1 - 9\,h^{3} + 9\,h^{3}$$

$$\displaystyle 12\,h^{3} = 1$$

$$\displaystyle 12(\dfrac{1}{12})\,h^{3} = 1(\dfrac{1}{12})$$

$$\displaystyle h^{3} = \dfrac{1}{12}$$

$$\displaystyle (h^{3})^{1/3} = (\dfrac{1}{12})^{1/3}$$

$$\displaystyle h = 0.4367902324$$ (one root)

Check

$$\displaystyle 15 + 3\,(0.4367902324)^{3} = 16 - 9\,(0.4367902324)^{3}$$

$$\displaystyle 15.25 \approx 15.25$$ (one root)

Yes

Find I. There should be three answers (roots).

$$\displaystyle 19 + 50\,i^{3} = 2 - \dfrac{4}{5}\,i^{3}$$

Last edited:

#### MarkFL

##### La Villa Strangiato
Staff member
Moderator
Math Helper
There are likely a complex conjugate pair of roots as well. A cubic equation (degree 3) will have 3 roots. Also, I wouldn't use the equal sign when subbing in a decimal approximations, but rather, I would use the LaTeX code:

\approx

#### TheJason

Staff member
Moderator
There are likely a complex conjugate pair of roots as well. A cubic equation (degree 3) will have 3 roots. Also, I wouldn't use the equal sign when subbing in a decimal approximations, but rather, I would use the LaTeX code:

\approx
Post edited. OK, we got one root at least. What would be the other two?

#### MarkFL

##### La Villa Strangiato
Staff member
Moderator
Math Helper
You could write the equation as:

$$\displaystyle 12h^3-1=0$$

Or

$$\displaystyle \left(\sqrt[3]{12}h\right)^3-1^3=0$$

Now, factor as the difference of cubes, and equate the two factors to zero to find all 3 roots.

TheJason

#### TheJason

Staff member
Moderator
You could write the equation as:

$$\displaystyle 12h^3-1=0$$

Or

$$\displaystyle \left(\sqrt[3]{12}h\right)^3-1^3=0$$

Now, factor as the difference of cubes, and equate the two factors to zero to find all 3 roots.
So in all cases of the cubed stuff, do you always use the $$\displaystyle -1 = 0\,$$ part? Why are those numbers used in this example?

Staff member