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b) 8a3 + 4a2b - 2ab2 – b3 = (8a3 – b3 ) + (4a2b - 2ab2 )
= (2a – b)(4a2 + 2ab + b2) + 2ab(2a – b)
= (2a – b)( 4a2 + 2ab + b2 + 2ab) = (2a – b)(2a + b)2
(a-b)^2=(a-b)(a-b)=a^2-ab-ab+b^2=a^2-2ba+b^2
(a-b)(a+b)=a^2+ab-ab-b^2=a^2-b^2
(a+3)^3=(a+b)^2*(a+b)
=(a^2+2ab+b^2)(a+b)
=a^3+a^2b+2a^2b+2ab^2+b^2a+b^3
=a^3+3a^2b+3ab^2+b^3
d) (8a3 – 27b3) – 2a(4a2 – 9b2)
= (2a – 3b)(4a2 + 6ab + 9b2) – 2a(2a – 3b)(2a + 3b)
= (2a – 3b)(4a2 + 6ab + 9b2 – 4a2 – 6ab) = 9b2(2a – 3b)
\(9a^2b+6ab^2+b^3-6ab-2b^2\)
\(=b\left(9a^2+6ab+b^2-6a-2b\right)\)
\(=b\left[\left(3a+b\right)^2-2\left(3a+b\right)\right]\)
\(=b\left(3a+b\right)\left(3a+b-2\right)\)
\(=b\left(9a^2+6ab+b^2\right)-2b\left(3a+b\right)\)
\(=b\left(3a+b\right)^2-2b\left(3a+b\right)\)
\(=b\left(3a+b\right)\left(3a+b-2\right)\)
\(9a^2b+6ab^2+b^3-6ab-2b^2\)
\(=b\left(9a^2+6ab+b^2-6a-2b\right)\)
\(=b\left[\left(3a+b\right)^2-2\left(3a+b\right)\right]\)
\(=b\left(3a+b\right)\left(3a+b-2\right)\)
a) Áp dụng HĐT 5 thu được ( 2 a - 3 b ) 3 .
b) Ta có 8 x 3 + 12 x 2 y + 6 xy 2 + y 3 = ( 2 x + y ) 3 .
Áp dụng HĐT 7 với A = 2x + y; B = z
( 2 x + y ) 3 - z 3 = (2x + y - z)(4 x 2 + y 2 + z 2 + 4xy + 2xz + zy).
b)
Sửa đề: \(125a^3+75a^2+15a+1\)
Ta có: \(125a^3+75a^2+15a+1\)
\(=\left(5a\right)^3+3\cdot\left(5a\right)^2\cdot1+3\cdot5a\cdot1^2+1^3\)
\(=\left(5a+1\right)^3\)
c) Ta có: \(64-96a+48a^2-8a^3\)
\(=-\left(8a^3-48a^2+96a-64\right)\)
\(=-\left[\left(8a^3-64\right)-48a\left(a-2\right)\right]\)
\(=-\left[\left(2a-4\right)\left(4a^2+8a+16\right)-48a\left(a-2\right)\right]\)
\(=-\left[\left(a-2\right)\left(8a^2+16a+32-48a\right)\right]\)
\(=-\left(a-2\right)\left(8a^2-32a+32\right)\)
\(=-8\left(a-2\right)\left(a^2-4a+4\right)\)
\(=-8\left(a-2\right)^3\)