phân tích đa thức thành nhân tử:
\(a\left(b-c\right)^3+b\left(c-a\right)^3+c\left(a-b\right)^3\)
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S
4 tháng 10 2018
Đặt A là tên biểu thức; \(a+b-c=x;b+c-a=y;c+a-b=z\)
Khi đó \(x+y+z=a+b-c+b+c-a+c+a-b=a+b+c\)
=>\(A=\left(x+y+z\right)^3-x^3-y^3-z^3=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=x^3+y^3+3xy\left(x+y\right)+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
\(=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(c+a-b+a+b-c\right)\)
\(=3.2b.2c.2a=24abc\)
SM
4 tháng 9 2018
A = ( a + b + c )3 + ( a - b - c )3 + ( b - c - a )3 + ( c - a - b )3
= [ ( a + b ) + c ]3 + [ ( a - b ) - c ]3 + [ ( - c ) - ( a - b ) ] 3 + [ c - ( a + b ) ]3
= ( a + b )3 + 3.( a + b )2.c + 3.( a + b ).c2 + c3 + ( a - b )3 - 3.( a - b )2.c + 3.( a - b ).c2 - c3 + ( - c3 ) + 3.( a - b )2.c - 3.( a - b ).c2 -(a- b)3
+ c3 + 3.( a + b )2.c - 3.( a + b ).c2 - ( a + b )3
= 6.( a + b )2 .c
NA
1
30 tháng 9 2018
\(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\))
\(=a^3\left(b-c\right)-b^3\left(a-c\right)+c^3\left(a-b\right)\)
\(=a^3\left(b-c\right)-b^3\left[\left(b-c\right)+\left(a-b\right)\right]+c^3\left(a-b\right)\)
\(=a^3\left(b-c\right)-b^3\left(b-c\right)-b^3\left(a-b\right)+c^3\left(a-b\right)\)
\(=\left(b-c\right)\left(a^3-b^3\right)-\left(a-b\right)\left(b^3-c^3\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(a^2+ab+b^2\right)-\left(b^2+bc+c^2\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a^2-c^2+ab-bc\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(a-c\right)\left(a+c\right)+b\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)
NA
1
28 tháng 9 2018
\(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(=a^3\left(b-c\right)-b^3\left[a-b+b-c\right]+c^3\left(a-b\right)\)
\(=a^3\left(b-c\right)-b^3\left(a-b\right)-b^3\left(b-c\right)+c^3\left(a-b\right)\)
\(=\left(b-c\right)\left(a^3-b^3\right)-\left(a-b\right)\left(b^3-c^3\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a^2+ab+b^2-b^2-bc-c^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a^2+ab-bc-c^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(a-c\right)\left(a+c\right)+b\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)
IM
5 tháng 10 2016
\(=a^3c-a^3b+c^3b-c^3a+b^3\left(a-c\right)\)
\(=b\left(c^3-a^3\right)+ac\left(a^2-c^2\right)+b^3\left(a-c\right)\)
\(=b\left(c-a\right)\left(c^2+ab+a^2\right)+ac\left(a+c\right)\left(a-c\right)+b^3\left(a-c\right)\)
\(=ac\left(a+c\right)\left(a-c\right)+b^3\left(a-c\right)-b\left(a-c\right)\left(c^2+ac+a^2\right)\)
\(=\left(a-c\right)\left[ac\left(a-c\right)+b^3-b\left(c^2+ac+a^2\right)\right]\)
\(=\left(a-c\right)\left[a^2c-c^2a+b^3-bc^2-bac-ba^2\right]\)
\(=\left(a-c\right)\left[a^2c-c^2a+b^3-bc^2-bac-ba^2\right]\)
18 tháng 9 2018
(a+b+c)3−a3−b3−c3(a+b+c)3−a3−b3−c3
=a3+3a2(b+c)+3a(b+c)2+(b+c)3−a3−b3−c3=a3+3a2(b+c)+3a(b+c)2+(b+c)3−a3−b3−c3
=3(b+c)(a2+ab+ac)+b3+3b2c+3bc2+c3−b3−c3=3(b+c)(a2+ab+ac)+b3+3b2c+3bc2+c3−b3−c3
=3(b+c)(a2+ab+ac+bc)=3(b+c)(a2+ab+ac+bc)
=3(b+c)[a(a+b)+c(a+b)]=3(b+c)[a(a+b)+c(a+b)]
=3(b+c)(a+b)(a+c)
mình lớp 6 ko biết làm
a(b-c)3+b(c-a)3+c(a-b)3
=a(b-c)3-b[(a-b)+(b-c)]+c(a-b)3
=a(b-c)3-b[(a-b)3+3(a-b)2(b-c)+3(a-b)(b-c)2+(b-c)3]
+c(a-b)3
=a(b-c)3-b(a-b)3+3b(a-b)2(b-c)+3b(a-b)(b-c)2+b(b-c)3
+c(a-b)3
=(b-c)3(a-b)-(a-b)3(b-c)-3b(a-b)(b-c)(a-b+b-c)
=(b-c)3(a-b)-(a-b)3(b-c)-3b(a-b)(b-c)(a-c)
=(a-b)(b-c)[(b-c)2-(a-b)2-3b(a-c)]