Question

Find the cube roots of 125(cos 288° + i sin 288°)

Answer 1

Let r(cos O + i sin O)  be a cube root of 125(cos 288 + i sin 288)
then
r^3(cos O + i sin O)^3  =  125(cos 288 + i sin 28)

so r^3 = 125  and  cos 3O + i sin 3O  =  cos 288 + i sin 288

so r  = 5  and 3O = 288 + 360p and O = 96 +  120p

so one cube root is   5 (cos 96 + i sin 96)

Im a little rusty at this stuff Its been a long time.

Im not sure of the other 2 roots

sorry cant help you any more

Answer 2

Answer:

The cube roots are

        5(cos96+isin96), 5(cos216+isin216) and 5(cos336+isin336)

Step-by-step explanation:

125(cos 288° + i sin 288°) can be written as 5³(cos 288° + i sin 288°)

5³(cos 288° + i sin 288°) $=5^3e^{i288^0}$

Complex number

         $e^{i\theta }=e^{i(\theta +2\pi) }=e^{i(\theta +4\pi) }$

So

    $5^3e^{i288^0}=5^3e^{i(288^0+360^0)}=5^3e^{i(288^0+720^0)}$

Finding cube root

   $\left (5^3e^{i288^0} \right )^{\frac{1}{3}}=5e^\frac{i288^0}{3}=5e^{i96^0}=5(cos96+isin96)\\\\\left (5^3e^{i(288^0+360^0)} \right )^{\frac{1}{3}}=5e^\frac{i(288^0+360^0)}{3}=5e^{i216^0}=5(cos216+isin216)\\\\\left (5^3e^{i(288^0+720^0)} \right )^{\frac{1}{3}}=5e^\frac{i(288^0+720^0)}{3}=5e^{i3366^0}=5(cos336+isin336)$

So the cube roots are

        5(cos96+isin96), 5(cos216+isin216) and 5(cos336+isin336)