Question

❤${\bold{\purple{\small{Good Afternoon}}}}$❤

If
${z}^{2} + \frac{1}{ {z}^{2} } = 34$
Then find the value of
${z}^{3} + \frac{1}{ {z}^{3} }$
​

Answer 1

Answer:

198

Step-by-step explanation:

Given Equation is z² + 1/z² = 34

⇒ z² + (1/z²) = 36 - 2

⇒ z² + (1/z²) + 2 = 36

⇒ (z + 1/z)² = 36

⇒ (z + 1/z) = 6.

On cubing both sides, we get

⇒ (z + 1/z)³ = (6)³

⇒ z³ + 1/z³ + 3(z + 1/z) = 216

⇒ z³ + 1/z³ + 3(6) = 216

⇒ z³ + 1/z³ + 18 = 216

⇒ z³ + 1/z³ = 216 - 18

⇒ z³ + 1/z³ = 198

Hope it helps!

Answer 2

$\huge\bf\mathscr\pink{Your\: Answer}$

198

step-by-step explanation:

Given,

${z}^{2} + \frac{1}{{z}^{2}}=34$

Adding 2 on both sides we get,

=> ${z}^{2} + \frac{1}{ {z}^{2} } + 2= 34 + 2$

=> ${z}^{2} + \frac{1}{ {z}^{2} } + 2× z × 1/z = 36$

=> ${(z+1/z)}^{2}$ = 36

=> z + 1/z = 6 ................ (i)

Now,

cubing both sides,

we get,


${(z+1/z)}^{3}$ = ${6}^{3}$............ from (i)

=> ${z}^{3}+\frac{1}{{z}^{3}}$ + 3 × z × 1/z( z+1/z ) = 216

=> ${z}^{3} + \frac{1}{ {z}^{3} }$ + 3 × 6 = 216

=> ${z}^{3} + \frac{1}{ {z}^{3} }$ = 216 - 18

=> ${z}^{3} + \frac{1}{ {z}^{3} }$ = 198,