maths ......equation

Equation C) can be transformed into [(1/x) - 3 ]2=2[(1/x) -3]+ 2*3+2*21
ie.adding 2*3 and subtracting 2*3 at the RHS of the equation; value remains unchanged.C) now becomes[(1/x) - 3 ]2=2[(1/x) -3]+ 2*24. Let y=[(1/x) - 3 ]
C) becomes y2 = 2y +2* 24=2(y + 24)which is equation a).Factorizing a) we get (y-8)(y+6)=0 y=8 or -6 Substituting value of y to find value of x, we get for y=8, 8=[(1/x)-3], x=1/11
for y=-6, x=-1/3

(a)

y^2 = 2(y + 24).

y^2 - 2y - 48 = 0

(y - 8) (y + 6) = 0

y = 8 or -6

(b)

(1/x^2) - (6/x) + 9
= (1/ x^2) ( 1 - 6x + 9x^2)
= (1/x^2) (3x-1)^2
= [ (3x-1) / x ]^2
=(3 - 1/x)^2

P.S. (1/x-3)^2 also correct

(c)
你這個問題, 如果你沒有打錯equation的話, 用result a and b係解唔到的
因為

(1/x^2) - (6/x) + 9 = 2[(1/x) + 21 ]
(3 - 1/x)^2 = 2[(1/x) + 21 ] (from b)

>>> (3 - 1/) 唔等於 1/x
>>> 解唔到 (result 係用唔到的)

仲有, 你條equation 都唔match question a 的equation, 最後的term 是 +21 , 而不是+24, 更加冇可能用到question a既result 啦
其實straight forward 就答到

(3 - 1/x)^2 = 2[(1/x) + 21 ]
(3 - 1/x)^2 = 2x ( 1+ 21x)
1 - 6x + 9x^2 = 2x + 42x^2
33x^2 + 8x - 1 = 0
(11x - 1) ( 3x + 1) = 0
x = 1/11 or -1/3