\(=a^3-3a^2+7a^2-21a-\left(8a-24\right)\)hay 

\(=a^2\left(a-3\right)+8a\left(a-3\right)-8\left(a-3\right)\)

\(=\left(a-3\right)\left(a^2+8a-8\right)\)

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\(a^3+4a^2-29a+24\)

\(=\left(a^3-3a^2\right)+\left(7a^2-21a\right)+\left(-8a+24\right)\)

\(=\left(a-3\right)\left(a^2+7a-8\right)\)

\(=\left(a-3\right)\left[\left(a^2-a\right)+\left(8a-8\right)\right]\)

\(=\left(a-3\right)\left(a-1\right)\left(a+8\right)\)

   \(a^3+4a^2-29a+24\)

= \(a^3-a^2+5a^2-5a-24a+24\)

= \(a^2\left(a-1\right)+5a\left(a-1\right)-24\left(a-1\right)\)

= \(\left(a-1\right)\left(a^2-3a+8a-24\right)\)

= \(\left(a-1\right)\left(a-3\right)\left(a+8\right)\)

a) \(a^3+4a^2-29a+24=\left(a^3-a^2\right)+\left(5a^2-5a\right)+\left(-24a+24\right)\)

\(=\left(a-1\right)\left(a^2+5a-24\right)=\left(a-1\right)\left(a^2+8a-3a-24\right)=\left(a-1\right)\left(a+8\right)\left(a-3\right)\)

b) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

Ta có \(\left(a+b+c\right)^3=a^3+b^3+c^3+3a^2b+3ab^2+3ac^2+3bc^2+3a^2c+3b^2c+6abc\)

\(\Rightarrow\left(a+b+c\right)^3-a^3-b^3-c^3=3a^2b+3ab^2+3ac^2+3bc^2+3a^2c+3b^2c+6abc\)

\(=3\left(a^2b+ab^2\right)+3\left(bc^2+ac^2\right)+3\left(a^2c+abc\right)+3\left(bc^2+abc\right)\)

\(=3\left(a+b\right)\left(ab+bc+ac+bc\right)=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

c) Theo trên ta có 

\(a^3+b^3+c^3-3abc=\left(a+b+c\right)^3-3\left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+3abc\right)\)

\(=\left(a+b+c\right)^3-3\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ca-3ab-3bc-3ca\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

d) \(x^5+x-1=\left(x^5-x^4+x^3\right)+\left(x^4-x^3+x^2\right)-\left(x^2-x+1\right)\)

\(=\left(x^2-x+1\right)\left(x^3+x^2-1\right)\)

E = x^3 - 3x^2 + 7x^2 - 21x - 8x + 24

  = x^2 ( x- 3 ) + 7x ( x- 3 ) - 8 ( x- 3 )

= ( x- 3 )(x^2 + 7x - 8 )

= ( x- 3 )[ x^2 + 8x - x - 8 ) 

= ( x -3 ) [ x(x + 8 ) - ( x + 8 ) ]

= ( x- 3  )( x - 1 )( x +  8)  

=a³-3a²+7a²-21a-8a+24

=a²(a-3)+7a(a-3)-8(a-3)

=(a-3)(a²+7a-8)

=(a-3)(a²-a+8a-8)

=(a-3)(a+8)(a-1)

khó quá nhỉ , toán 1 mà

a) \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)+b^2\left[\left(c-b\right)-\left(a-b\right)\right]+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)

\(=\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(a+b-b-c\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(a-c\right)\)

c) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\left(1\right)\)

Đặt \(x^2+8x+11=y\)Thay vào (1) ta được 

\(\left(y-4\right)\left(y+4\right)+15\)

\(=y^2-16+15\)

\(=y^2-1\)

\(=\left(y-1\right)\left(y+1\right)\)

\(=\left(x^2+8x+10\right)\left(x^2+8x+11\right)\)

nhầm dong cuối là 

\(\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

toàn mấy baiif đơn giản trong sách giao khoa mà, bạn tự suy nghĩ đi chứ . Mình nghĩ thay vì viết hàng loạt những bài dài thế này thì bạn nên dành thời gian để suy nghĩ thì hơn đó bạn

TỚ CHỈ LÀM ĐG CÂU 3 THÔI

24 = 2^3.3

84 = 2^2.3.5

CÂU 3 AN LÀM ĐÚNG