Question

if zsquare + 1 upon z square is equal to 18 find the value of Z cube minus 1 upon z cube using only positive value of that minus 1 by z​

Answer 1

Answer:

52

Step-by-step explanation:

z²+1/z²=18

subtracting 2 both sides

z²-2+1/z²=16

z²-2*z*1/z+1/z²=16

(z-1/z)²=16

z-1/z=±4

================================

taking z-1/z=4

Now z³-1/z³=(z-1/z)³+3*z*1/z(z-1/z)

=(4)³-3*1*(4)

=64-12

=52

Answer 2

$\huge\mathfrak{\underline{\underline{QuestioN:}}}$

If $\sf{(z^2)+}$$\sf\frac{(1)}{(z^2)}=18$, find the value of $\sf{(z^3)-}$$\sf\frac{(1)}{(z^3)}$, using only the positive value of $\sf{z-}$$\sf\frac{1}{z}$.

$\huge\mathfrak\red{\underline{\underline{AnsweR:}}}$

Given:

$\sf{z^2}$ + $\sf\frac{1}{z^2}=18$

$\implies$$\sf{z^2}$ + $\sf\frac{1}{z^2}-2=18-2$

$\implies$$\sf{z^2}$ + $\sf\frac{1}{z^2}-2.z.$$\sf\frac{1}{2}=16$

$\implies$$\sf{(z)-}$$\sf\frac{(1)}{z}^2=16=(4)^2$

$\implies$$\sf{z-}$$\sf\frac{1}{z}=4$

Now, $\sf{z^3}$-$\sf\frac{1}{z^3}=$$\sf{(z)-}$$\sf\frac{(1)}{(z)}$$\sf{(z^2+z.}$$\sf\frac{1}{z}+{1}{z^2)}$

= $\sf{z-}$$\sf\frac{1}{z}$$\times$

$\sf{z^2+1+}$$\sf\frac{1}{z^2}$

= $\sf{z-}$$\sf\frac{1}{z}=4$$\times$$\sf{z^2+}$$\sf\frac{1}{z^2}-2+3$

By solving it we get,

=(4)[(4)²-3]

= (4)(16+3)

= (4)(19)

= 76.