Question

If X square + 1 upon x square is equal to 7 and X is not equal to zero find the value of 7 x cube + 8 x minus 7. X cube minus 8 x

Answer 1

Bro,

7x³ + 8x - 7x³ - 8x

this eqñ will give you 0 as an answer

(7x³-7x³) + (8x-8x) = 0 +0 = 0

Answer 2

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

65√5 is the correct answer

Please make me brainliest