Phân tích đa thức thành nhân tử :
a ( b - c )3 + b ( c - a )3 + c ( a + b )3
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TM
0
HT
0
G
1
31 tháng 8 2019
\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)
\(=\left(ab^3-a^3b\right)+\left(bc^3-ac^3\right)+\left(a^3c-b^3c\right)\)
\(=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)\)
\(=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c-abc+b^2c\right)\)
BD
0
24 tháng 10 2021
Đặt \(\left\{{}\begin{matrix}a+b-c=x\\b+c-a=y\\c+a-b=z\end{matrix}\right.\Leftrightarrow x+y+z=a+b+c\)
Do đó \(A=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(\Leftrightarrow A=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)-x^3-y^3-z^3\\ \Leftrightarrow A=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(\Leftrightarrow A=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(c+a-b+a+b-c\right)\\ \Leftrightarrow A=3\cdot2b\cdot2c\cdot2a=24abc\)
AN
2
8 tháng 7 2021
a3(b−c)+b3(c−a)+c3(a−b)
=a3b−a3c+b3c−b3a+c3(a−b)
=(a3b−b3a)−(a3c−b3c)+c3(a−b)
=ab(a2−b2)−c(a3−b3)+c3(a−b)
=ab(a−b)(a+b)−c(a−b)(a2+ab+b2)+c3(a−b)
=(a−b)[ab(a+b)−c(a2+ab+b2)+c3]
=(a−b)(a2b+ab2−a2c−abc−b2c+c3)
=(a−b)[(a2b−a2c)+(ab2−abc)−(b2c−c3)]
=(a−b)[a2(b−c)+ab(b−c)−c(b2−c2)]
=(a−b)[a2(b−c)+ab(b−c)−c(b−c)(b+c)]
=(a−b)(b−c)[a2+ab−c(b+c)]
=(a−b)(b−c)(a2+ab−bc−c2)
=(a−b)(b−c)[(a−c)(a+c)+b(a−c)]
=(a−b)(b−c)(a−c)(a+b+c)
NT
0
D
2
S
13 tháng 8 2018
=a3(b-c)-b3(a-c)+c3(a-c-b+c)
=a3(b-c)-b3(a-c)+c3(a-c)-c3(b-c)
=(a3-c3)(b-c)-(b3-c3)(a-c)
=(a-c)(a2+ac+c2)(b-c)-(b-c)(b2+bc+c2)(a-c)
=(b-c)(a-c)(a2+ac+c2-b2-bc-c2)
=(b-c)(a-c)[(a-b)(a+b)+c(a-b)]
=(a-b)(b-c)(a-c)(a+b+c)
D
3 tháng 9 2018
Ta có:\(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(=a^3\left(b-c\right)-b^3\left(a-c\right)+c^3\left(a-c-b+c\right)\)
\(=a^3\left(b-c\right)-b^3\left(a-c\right)+c^3\left(a-c\right)-c^3\left(b-c\right)\)
\(=\left(a^3-c^3\right)\left(b-c\right)-\left(b^3-c^3\right)\left(a-c\right)\)
\(=\left(a-c\right)\left(a^2+ac+c^2\right)\left(b-c\right)-\left(b-c\right)\left(b^2+bc+c^2\right)\left(a-c\right)\)
\(=\left(b-c\right)\left(a-c\right)\left(a^2+ac+c^2-b^2-bc-c^2\right)\)
\(=\left(b-c\right)\left(a-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)
DQ
0
DM
0
a ( b - c )3 + b ( c - a )3 + c ( a + b )3
= a ( b3 -3b2c+3bc2- c3 )+ b ( c3 -3c2a+3ca2- a3 ) + c ( a3 -3a2b+3ab2- b3 ) (khai triển ra)
= (ab3 - 3ab2c + 3abc2- ac3) + (bc3 - 3abc2 + 3a2bc- a3b) + (a3c -3a2bc+3ab2c- b3c) (nhân vào)
= ab3 - 3ab2c + 3abc2- ac3 + bc3 - 3abc2 + 3a2bc- a3b + a3c -3a2bc+3ab2c- b3c (phá ngoặc)
= ab3-ac3+bc3- a3b + a3c - b3c (triệt tiêu các hạng tử trái dấu)
=a2bc+ab2c+abc2+a3b+a2b2+a2bc-a3c-a2bc-a2c2+a2c2+abc2+ac3-a2b2-ab3-ab2c+ab2c+b3c+b2c2-abc2-b2c2-bc3-a2bc-ab2c-abc2
= (a2bc + ab2c + abc2) +(a3b + a2b2 + a2bc) - (a3c - a2bc - a2c2) +(a2c2 + abc2 +ac3) -
(a2b2 + ab3 + ab2c) + (ab2c + b3c + b2c2) - (abc2 + b2c2 + bc3) - (a2bc + ab2c + abc2)
= abc(a + b + c) +a2b(a + b + c) - a2c(a + b + c) + ac2(a + b + c) - ab2(a + b + c) + b2c(a + b + c) - bc2(a + b + c) - abc(a + b+ c)
= (a +b +c)(abc + a2b - a2c + ac2 - ab2 + b2c - bc2 - abc)
= (a + b+ c) [(a2b - abc)+(abc - bc2) - (a2c - ac2) - (ab2 - b2c)]
= (a + b + c) [ab(a - c) + bc(a - c) - ac(a - c) - b2(a - c)]
= (a + b + c)(a - c)(ab + bc - ac - b2)
= (a +b + c)(a - c) [(ab - ac) - (b2 - bc)]
= (a + b+ c)(a - c) [a(b - c) - b(b - c)]
= (a + b + c)(a - c)(b - c)(a - b)
A = (b - c)³ + (c - a)³ + (a - b)³
Áp dụng hằng đẳng thức : a³ + b³ = (a + b)³ - 3ab(a + b) :
A = [(b - c)³ + (c - a)³] + (a - b)³
= [(b - c) + (c - a)]³ - 3(b - c)(c - a)[(b - c) + (c - a)] + (a - b)³
= (b - a)³ - 3(b - c)(c - a)(b - a) + (a - b)³
= [- (a - b)³] - 3(b - c)(c - a)[- (a - b)] + (a - b)³
= - (a - b)³ + 3(a - b)(b - c)(c - a) + (a - b)³
= 3(a - b)(b - c)(c - a)