Question

Please help me. ..........

Answer 1

Answer:

±756

Step-by-step explanation:

Given :

$x^2 + \frac{1}{x^2} = 83$

Then,

Find : $x^3-\frac{1}{x^3}$

Solution :

We know that,

a³ - b³ = (a - b) (a² - ab + b²)

__

$x^2 + \frac{1}{x^2} = 83$

By subtracting by 2 , both the sides,

We get,

⇒ $x^2 +\frac{1}{x^2} - 2 = 83 - 2$

⇒ $x^2 +\frac{1}{x^2} - 2(x)(\frac{1}{x}) = 83 - 2$

⇒ $(x-\frac{1}{x})^2 = 81 = 9^2$

⇒ $x-\frac{1}{x} = 9$ (or) $-9$

___

⇒ $(x)^3 - (\frac{1}{x})^3 = x^3 - \frac{1}{x^3} = (x - \frac{1}{x})(x^2 + \frac{1}{x^2} + (x)(\frac{1}{x} )) = 9 (83 + 1) = 9 \times 84 = 756$ (or) $-9 (83 + 1) = -9 \times 84 = -756$

Answer 2

Step-by-step explanation:

x^2+1/x^2=83.

we know that (x-1/x)^2=x^2+(1/x)^2-2.

=83-2

. =81.

(x-1/x) =9.

now...

x^3-1/x^3=(x-1/x)^3+3(x-1/x ).

=9^3+3*9

. =729+27

. =756.

I hope this is right and helpful for u.

please follow me.