Question

if x^2 +1/x^2 = 7 , find the value of : 7x^3 +8x -7/x^3 - 8/x
${ques 9}$

Answer 1

7(x^3 - 1/x^3) + 8(x - 1/x)
Since a^3 - b^3 = (a-b)(a^2 + ab + b^2)
7[(x-1/x)(x^2 + 1 + 1/x^2)] + 8(x-1/x)
7[(x-1/x)(7+1)] + 8(x-1/x)
56(x-1/x) + 8(x-1/x)
(x-1/x)(56+8)
64(x-1/x)
Since (x-1/x)^2=x^2 + 1/x^2 - 2=7-2=5
Therefore x-1/x =√5
64√5

Answer 2

Answer:

Given- x^2 + 1/x^2 = 7

Find- 7x^3 + 8x - 7/x^3 - 8/x

Solution =

7x^3 - 7/ x^3 + 8x - 8/x

Now,

7 ( x^3- 1/x^3) + 8 (x - 1/x)

Since,

x^3+- 1/ x^3 = ( x-1/x ) ( x^2+1/x^2+1)

7 { ( x-1/x )( x^2+1/x^2+1 )} + 8 ( x-1/x )

7 {(x -1/x ) ( 7+1 )} +8 ( x- 1/x )

56( x-1/x ) + 8 ( x-1/x )

(56+8) ( x-1/x ) ^2

64 ( x^2 + 1/x^2 -2)

64 (7-2)

64√5

Step-by-step explanation: