Question

IF z ^ 2 + 1/(z ^ 2) = 6 find the value of z ^ 3 - 1/(z ^ 3) using only the positive value of z - 1/z​

Answer 1

Let that,

$\rm{z-\dfrac{1}{z}=x}$

We know the formula of,

$\cdots\longrightarrow\boxed{\rm{A^{2}-2AB+B^{2}=(A-B)^{2}}}$

Here,

• $\rm{A=z}$
• $\rm{B=\dfrac{1}{z}}$

$\rm{z^{2}-2\cdot z\cdot \dfrac{1}{z}+\dfrac{1}{z^{2}}=x^{2}}}$

$\rm{z^{2}-2+\dfrac{1}{z^{2}}=x^{2}}$

$\rm{x^{2}=6-2}$

$\rm{x^{2}-4=0}$

$\therefore \rm{x=2}}$

Hence,

$z-\dfrac{1}{z}=2$

We know the formula of,

$\cdots\longrightarrow\boxed{\rm{A^{3}-B^{3}=(A-B)^{3}+3AB(A-B)}}$

Here,

• $\rm{A=z}$
• $\rm{B=\dfrac{1}{z}}$

$\rm{\left(z-\dfrac{1}{z}\right)^{3}+3\cdot z\cdot \dfrac{1}{z}\cdot \left(z-\dfrac{1}{z} \right)=2^{3}+3\cdot2}$

$\rm{z^{3}-\dfrac{1}{z^{3}}=8+6}$

$\therefore \rm{z^{3}-\dfrac{1}{z^{3}}=14}$

Hence shown that

$\cdots\longrightarrow\boxed{\rm{\rm{z^{3}-\dfrac{1}{z^{3}}=14}}}$