Question

if z^2 +1/z^2 =14 find the value of z^3 + 1/z^3

Answer 1

Hi there!
( Z+1/Z)^2=14

Z2+1/Z 2= 14 +2=16=4^2

Z+1/Z=4

(Z+1/Z)3=64

Z3+1/z3+3(z+1/z)=64

Z3+1/z3+12=64

Z3+1/z3=64-12=52

Thus the value of z3+1/z3=52

Hope this helps you!

Answer 2

Given that,

The value of $z^2+\dfrac{1}{z^2}=14$

To find,

The value of $z^3+\dfrac{1}{z^3}$.

Solution,

Firstly we can find the value of $(z+\dfrac{1}{z})^2$.

$(z+\dfrac{1}{z})^2=z^2+\dfrac{1}{z^2}+2z\times \dfrac{1}{z}\\\\=z^2+\dfrac{1}{z^2}+2$

Since,  $z^2+\dfrac{1}{z^2}=14$

So,

$(z+\dfrac{1}{z})^2=14+2\\\\=16$    

or

$(z+\dfrac{1}{z})=4$ ....(1)

Cubing both sides,

$(z+\dfrac{1}{z})^3=4^3$

We know that, $(a+b)^3=a^3+b^3+3a^2b+3ab^2$

So,

$(z+\dfrac{1}{z})^3=4^3\\\\z^3+\dfrac{1}{z^3}+3z^2\times \dfrac{1}{z}+3z\times \dfrac{1}{z^2}=64\\\\z^3+\dfrac{1}{z^3}+3z+\dfrac{3}{z}=64\\\\z^3+\dfrac{1}{z^3}+3(z+\dfrac{1}{z})=64$

From equation (1).

$z^3+\dfrac{1}{z^3}+3\times 4=64\\\\z^3+\dfrac{1}{z^3}=52$

So, the value of $z^3+\dfrac{1}{z^3}$ is 52.