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3 ( a+b)(a+c)(b+c)
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23 tháng 7 2021
Ta có: \(\left(b-c\right)^3+\left(c-a\right)^3-\left(a-b\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(b-c+c-a\right)\left[\left(b-c\right)^2-\left(b-c\right)\left(c-a\right)+\left(c-a\right)^2\right]-\left(a-b\right)\left[1+3\left(b-c\right)\left(c-a\right)\right]\)
\(=\left(b-a\right)\left(b^2-3bc+3c^2+ab-3ac+a^2\right)-\left(a-b\right)\left(1+3bc-3ab-3c^2+3ac\right)\)
\(=\left(b-a\right)\left(b^2-3bc+3c^2+ab-3ac+a^2+1+3bc-3ab-3c^2+3ac\right)\)
\(=\left(b-a\right)\left(b^2-2ab+a^2+1\right)\)
\(=\left(b-a\right)^3+\left(b-a\right)\)
\(=b^3-3b^2a+3ba^2-a^3+b-a\)
NH
1
3 tháng 8 2016
\(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Đặt a+b=x ; b+c=y; c+a=z ta có:
\(x^3+y^3+z^3-3xyz\)
=\(\left(x+y\right)^3-3x^2y-3xy^2+z^3-3xyz\)
=\(\left[\left(x+y\right)^3+z^3\right]-\left(3x^2y+3xy^2+3xyz\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
xong thay vào
NP
2
12 tháng 7 2015
Đặt a+b = x ; b+c = y ; c+a = z
=> H = x^3 +y^3 +z^3 -3.x.y.z
= [x+y]^3 -3.x^2.y -3.x.y^2+ z^3 - 3.x.y.z
= {[x+y]^3+z^3} -3.x.y[x+y+z]
= [x+y+z].{[x+y]^2-[x+y].z+z^2} +3.x.y[x+y+z]
= [x+y+z] . [x^2+y^2+2.x.y-x.z-y.z+z^2+3.x.y]
= [x+y+z]. [x^2+y^2+z^2-xy-y.z-x.z]
= [a+b+b+c+c+a]. {[a+b]^2+[b+c]^2+[c+a]^2-[a+b].[b+c]-[a+b].[a+c] - [b+c].[c+a]}
= 2.[a+b+c] .[a^2+b^2 +b^2 +c^2 +c^2 +a^2 +2.ab.+2.bc+2.ac-ab-b^2-ac-bc-a^2-ab-ac-bc-bc-c^2-ab-ac]
= 2.[a+b+c].[a^2+b^2+c^2-ab-ac-bc]
NH
0
TD
2
TM
1
ML
3 tháng 7 2015
Áp dụng hằng đẳng thức dưới dạng
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)
\(\left(a+b+c\right)^3+\left(a-b-c\right)^3=\left(2a\right)^3-3\left(a+b+c\right)\left(a-b-c\right).2a\)
\(\left(b-c-a\right)^3+\left(c-a-b\right)^3=\left(-2a\right)^3-3\left(b-c-a\right)\left(c-a-b\right).\left(-2a\right)\)
\(\Rightarrow\left(a+b+c\right)^3+\left(a-b-c\right)^3+\left(b-c-a\right)^3+\left(c-a-b\right)^3\)
\(=\left(2\right)^3+\left(-2a\right)^3-6a\left[a+\left(b+c\right)\right]\left[a-\left(b+c\right)\right]+6a\left[-a+\left(b-c\right)\right]\left[-a-\left(b-c\right)\right]\)
\(=-6a\left\{a^2-\left(b+c\right)^2-\left[\left(-a\right)^2-\left(b-c\right)^2\right]\right\}\)
\(=-6a\left\{a^2-a^2+\left(b-c\right)^2-\left(b+c\right)^2\right\}\)
\(=-6a\left[b-c+b+c\right]\left[b-c-\left(b+c\right)\right]=-6a.2b.\left(-2c\right)\)
\(=24abc\)
PT
0
`#3107.101107`
`3(a + b)(a + c)(b + c)`
`= 3(a^2 + 2ab + bc)(b + c)`
`= 3(a^2b + a^2c + 2ab^2 + 2abc + b^2c + bc^2)`
`= 3a^2b + 3a^2c + 6ab^2 + 6abc + 3b^2c + 3bc^2`