Given z²+1/z² =14
using above equation,
z²+1/z² = 14
add 2 on both side,
z² + 1/z² + 2 = 14 + 2
the LHS is whole square of z + 1/z
so we can write it
(z + 1/z)² = 16z + 1/z = 4
Now we know,
(a + b)³ = a³ + b³ + 3ab(a+b)
here, a = z and b = 1/z,
substituting a and b in equation we will get,
(z + 1/z)³ = z³ + 1/z³ + 3z/z( z+ 1/z)
4³ = z³ + 1/z³ + 3×4
z³+ 1/z³ = 64-12
z³ + 1/z³ = 52
Answer:
52.
Step-by-step explanation:
Using the identity a^3 + b^3 = (a + b)(a^2 - ab + b^3):
z^3 + 1/z^3 = (z + 1/z)(z^2 - z *1/z + 1/z^2)
= (z + 1/z) (z^2 - 1 + 1/z^2)
= (z + 1/z) (14 - 1 )
= 13(z + 1/z).
Now using the identity a^2 + b^2 = (a + b)^2 - 2ab:
z^2 + 1/z^2 = ( z + 1/z)^2 - 2(z * 1/z) = 14 So:
( z + 1/z)^2 - 2 = 14
( z + 1/z)^2 = 16
Taking z + 1/z = 4 we therefore have:
z^3 + 1/z^3 = 13 * 4 = 52.
