Answered

solve 64x^3 +1 =0

the back of the texts books gives me the answers -1/4 and 1 plus or minus i√3 all over 8

I understand where the -1/4 comes from . I just don't understand the second answer. I know they used the quadratic formula to get it but I always end up with something to totally different

Verified Answer :

naǫ

Expand the expression using the formula:
\(a^3+b^3=(a+b)(a^2 - ab + b^2)\)

\(64x^3+1=0 \\ (4x)^3+1^3=0 \\ (4x+1)((4x)^2- 4x \times 1+1^2)=0 \\ (4x+1)(16x^2-4x+1)=0 \\ 4x+1=0 \ \lor \ 16x^2-4x+1=0\)

The first equation:
\(4x+1=0 \\ 4x=-1 \\ x=-\frac{1}{4}\)

The second equation:
\(16x^2-4x+1=0 \\ \\ a=16 \\ b=-4 \\ c=1 \\ b^2-4ac=(-4)^2-4 \times 16 \times 1=16-64=-48 \\ \sqrt{b^2-4ac}=\sqrt{-48}=\sqrt{48} \times \sqrt{-1}=\sqrt{16 \times 3} \times \sqrt{-1}=\pm 4\sqrt{3} i \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-4) \pm 4\sqrt{3}i}{2 \times 16}=\frac{4 \pm 4\sqrt{3}i}{32}=\frac{4(1 \pm \sqrt{3}i)}{4 \times 8}=\frac{1 \pm \sqrt{3}i}{8} \\ x=\frac{1-\sqrt{3}i}{8} \ \lor \ x=\frac{1+\sqrt{3}i}{8}\)

The final answer:
\(\boxed{x=-\frac{1}{4} \hbox{ or } x=\frac{1-\sqrt{3}i}{8} \hbox{ or } x=\frac{1+\sqrt{3}i}{8}}\)