4) $x^{6} + 65 x^{3} + 64 = 0 (4)$
A) ${- 4, - 2 + i \sqrt{11}, - 2 - i \sqrt{11}, - 1, \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}}$
B) ${- 4, \frac{- 1 + i \sqrt{47}}{3}, \frac{- 1 - i \sqrt{47}}{3}, - 1, \frac{1 + i \sqrt{11}}{2}, \frac{1 - i \sqrt{1}}{2}$
C) ${0, - 4, 2 + 2 i \sqrt{3}, 2 - 2 i \sqrt{3}, - 1, 1}$
D) ${- 4, 2 + 2 i \sqrt{3}, 2 - 2 i \sqrt{3}, - 1, \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}}$
$- 1 -$

Answer

Expert–verified

Hide Steps

Answer

Solve for $u$ and then for $x$ : $u = - 1, - 64$ which yields $x = {- 4, 2 + 2 i \sqrt{3}, 2 - 2 i \sqrt{3}, - 1, \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}}$

Steps

Step 1 ：Let $u = x^{3}$ , then $u^{2} + 65 u + 64 = 0$

Step 2 ：Factor the quadratic: $(u + 1) (u + 64) = 0$

Step 3 ：Solve for $u$ and then for $x$ : $u = - 1, - 64$ which yields $x = {- 4, 2 + 2 i \sqrt{3}, 2 - 2 i \sqrt{3}, - 1, \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}}$

4) x6+65x3+64=0(4)A) {−4,−2+i11,−2−i11,−1,1+i32,1−i32}B) {−4,−1+i473,−1−i473,−1,1+i112,1−i12C) {0,−4,2+2i3,2−2i3,−1,1}D) {−4,2+2i3,2−2i3,−1,1+i32,1−i32}−1−

Answer

Popular Problems