I thought this same question was asked and answered already before?

Are you not satisfied with the answer?

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Okay, let me pitch in.

You were okay until the

"Then I divide by (x-1) to get x^2 + ax + 9 = 0"

What you did is you factored

(x^3 - x^2) + a(x^2 - x) + 9(x - 1) = 0

into

(x-1)(x^2 +ax +9) = 0

What that means now is that (x-1) is a factor of the original

x^3 + (a-1)x^2 + (9-a)x - 9 = 0 ----(1)

So,

x-1 = 0

x = 1 <----this x-value makes (1) true.

Let us check, substitute that into (1),

1^3 +(a-1)(1^2) +(9-a)(1) -9 = 0

1 +(a-1) +(9-a) -9 = 0

1 +a -1 +9 -a -9 = 0

0 = 0

See, it is true.

Then, the other factor is

x^2 +ax +9 = 0 ---------(2)

We know x=1 is okay, so, plugging it in,

1^2 +a*1 +9 = 0

1 +a +9 = 0

a = -10 --------answer?

We check that.

Plug that into (2),

x^2 +ax +9 = 0 ---------(2)

x^2 -10x +9 = 0

(x-9)(x-1) = 0

x = 9 or 1 --------two real numbers that satisfy (2).

Will those satisfy (1) also?

Should be.

Check,

>>>When x=9, and a = -10,

x^3 + (a-1)x^2 + (9-a)x - 9 = 0 ----(1)

9^3 +(-10 -1)(9^2) +(9 +10)(9) -9 =? 0

729 -11(81) +19(9) -9 =? 0

0 =? 0

Yes, so, OK.

>>>when x=1, and a = -10,

x^3 + (a-1)x^2 + (9-a)x - 9 = 0 ----(1)

1^3 +(-10 -1)(1^2) +(9 +10)(1) -9 =? 0

1 -11(1) +19(1) -9 =? 0

0 =? 0

Yes, so, OK.

Therefore, a = -10 --------answer.