Question

If $\rm x^2+\dfrac{1}{x^2}=7 \ and \ x \neq 0,$ find the value of

$\rm 7x^3+8x-\dfrac{7}{x^3} - \dfrac{8}{x}$

please fast​

Answer 1

x² + 1/x² = 7

x² + 1/x² - 2 = 5

(x-1/x)² = 5

x - 1/x = root 5

x³ - 1/x³ = (x-1/x) (x² + 1/x² + 1)

x³ - 1/x³ = (root 5)( 7+1)

x³ - 1/x³ = 8 root 5

7x³ + 8x - 7/x³ - 8/x

7x³ - 7/x³ + 8x - 8/x

7( x³ - 1/x³) + 8( x- 1/x) ( putting the value of x³-1/x³ and x-1/x)

7( 8 root 5) + 8( root 5)

(8 root 5) (7+ 1)

64 root 5

Answer 2

Answer:

-(a+b+c)

Step-by-step explanation:

Given :  

To Find : x

Solution :

Hence The value of x is -(a+b+c