不明白中三的因式分解...

You should look at this word carefully below:
"由這"
That means you MUST use Q.(a) provided.

We have known that:
x³ + 27 = (x+3)(x²-3x+9)

Take some common factors (6&27,the HCF is 3) and
submit in the next question.
x³ - 7x + 6
Method:
1.You use this formula
x³ + 27=(x+3)(x²-3x+9)
2.Sub. in question x³ - 7x + 6
x³ - 7x + 6
=x³ - 7x + 6 - 27+ 27
=x³ - 7x - 21+ 27
=x³ +27 - 7x - 21
=(x+3)(x²-3x+9)-7(x+3)
=(x+3)(x²-3x+9-7)
=(x+3)(x²-3x+2)
After cross multiplication
=(x+3)(x-1)(x-2)
=(x-1)(x-2)(x+3)

OR
x³ - 7x + 6
=x³ - 1³ - 7x + 7
=(x-1)(x²+x+1) - 7x + 7
=(x-1)(x²+x+1) - 7 (x -1)
=(x-1)(x²+x+1-7)
=(x-1)(x²+x-6)
After cross multiplication
=(x-1)(x-2)(x+3)

In fact that there are 7 types to factorize.
1.x³ - 7x + 6
=x³ - 4x - 3x+ 6
2.x³ - 7x+ 6
= - 7x³ - 7x - 6x³+6
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etc.

You can use the answer to cal. the original question.
After you cal. it,the ans is right.

For (a), do you remember this identity?

x^3 + y^3 = (x + y) (x^2 - xy + y^2)

x^3 + 27
= x^3 + 3^3
= (x+3)(x^2-3x+9)

b)
x^3-7x+6
= x^3 + 27 - 7x - 21
= (x^3 + 27) - 7*(x + 3)
= (x+3)(x^2-3x+9) - 7*(x+3)
= (x+3)[(x^2-3x+9)-7]
= (x+3)(x^2-3x+2)
= (x+3)(x-1)(x-2)