Given:
[tex]a^2+3a+9=0[/tex]
If we need to find what [tex]a^3[/tex] is, we can solve for a and plug it in. Let's use the quadratic formula to solve for a. The quadratic formula is:
[tex]a=\frac{-b+\sqrt{b^2-4ac}}{2a}[/tex] and [tex]a=\frac{-b-\sqrt{b^2-4ac}}{2a}[/tex]
Let's identify our a, b, and c:
a: 1
b: 3
c: 9
Plug in the values into the quadratic formula:
[tex]a=\frac{-3+\sqrt{(3)^2-4(1)(9)}}{2(1)}[/tex]
Simplify everything under the radical:
[tex]a=\frac{-3+\sqrt{-27}}{2}[/tex]
Simplify the radical:
[tex]a=\frac{-3+3i\sqrt{3}}{2}[/tex]
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Let's solve for the second part of the quadratic formula. Everything will be the same, so we can just replace the + with a -.
[tex]a=\frac{-3+3i\sqrt{3}}{2}[/tex] and [tex]a=\frac{-3-3i\sqrt{3}}{2}[/tex]
These two are your answers. Now, since we know what a is, let's cube this value:
[tex](a=\frac{-3+3i\sqrt{3}}{2})^3[/tex]
Simplify:
[tex]27[/tex]
You will get the same answer if you cube both of the quadratic answers. So, your final answer:
[tex]a^3=27[/tex]
