Question

\(\sf Z^2+\frac{1}{Z^2}=11\), find \(\sf Z^3-\frac{1}{Z^3}\)

Answer 1

@Astrophysics @ParthKohli

Answer 2

\[Z^2+\frac{ 1 }{ Z^2 } = 11\]find\[Z^3-\frac{ 1 }{ Z^3 }\] latex never works for me on the original post

Answer 3

\[a^3 - b^3 = (a-b) (a^2 + ab + b^2)\]\[z^3 - \frac{1}{z^3} = \left(z - 1/z\right)(z^2 + 1/z^2 + 1)\]\[z^2 + \frac{1}{z^2}-2 = 9\Rightarrow (z - 1/z)^2 = 3^2\]\[\Rightarrow z - 1/z = \pm 3\]

Answer 4

\[z^3 - 1/z^3 = \pm 3 \cdot (11+1)\]

Answer 5

Yeah dawg!

Answer 6

Thanks

Answer 7

no

Answer 8

Lol