Question

If z Square +1 upon z square is equal to 14 find the value of z cube plus one upon z cube

Answer 1

$\textbf{\huge{ANSWER:}}[tex] Given: [tex] {z}^{2} + \frac{1}{ {z}^{2} } = 14$

To find:

${z}^{3} + \frac{1}{ {z}^{3} }$

Solution:

$({z + \frac{1}{z} )}^{2} = {z}^{2} + \frac{1}{ {z}^{2} } + 2 \times z \times \frac{1}{z} \\ \\ = > ({z + \frac{1}{z} )}^{2} = 14 + 2 \\ \\ = > ({z + \frac{1}{z} )}^{2} = 16 \\ \\ = >( z + \frac{1}{z} ) = \sqrt{16} \\ \\ = > (z + \frac{1}{z} ) = 4 \\ \\ now \\ \\ = > { (z + \frac{1}{z} ) }^{3} = {z}^{3} + \frac{1}{ {z}^{3} } + 3 \times z \times \frac{1}{z } \\ \\ = > { (z + \frac{1}{z} ) }^{3} = {z}^{3} + \frac{1}{ {z}^{3} } + 3 \times 4 \\ \\ = > {4}^{3} = {z}^{3} + \frac{1}{ {z}^{3} } + 12 \\ \\ = > {z}^{3} + \frac{1}{ {z}^{3} } = 64 - 12 \\ \\ = > {z}^{3} + \frac{1}{ {z}^{3} } = 52$


Hope it Helps!! :)

Answer 2

hey there,!!!

so+1/z³ = 52. .

the whole solution given in attachment !!

thank you !!

Hope it helps.

and All the best !!