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a )
`VP= (a+b)^3-3ab(a+b)`
`=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2`
`=a^3+b^3 =VT (đpcm)`
b)
b) Ta có
`VT=a3+b3+c3−3abc`
`=(a+b)3−3ab(a+b)+c3−3abc`
`=[(a+b)3+c3]−3ab(a+b+c)`
`=(a+b+c)[(a+b)2+c2−c(a+b)]−3ab(a+b+c)`
`=(a+b+c)(a2+b2+2ab+c2−ac−bc−3ab)`
`=(a+b+c)(a2+b2+c2−ab−bc−ca)=VP`
a) Ta có:
`VP= (a+b)^3-3ab(a+b)`
`=a^3 + b^3+3ab ( a + b )- 3ab ( a + b )`
`=a^3 + b^3=VT(dpcm)`
b) Ta có
`VT=a^3+b^3+c^3−3abc`
`=(a+b)^3−3ab(a+b)+c^3−3abc`
`=[(a+b)^3+c^3]−3ab(a+b+c)`
`=(a+b+c)[(a+b)^2+c^2−c(a+b)]−3ab(a+b+c)`
`=(a+b+c)(a^2+b^2+2ab+c^2−ac−bc−3ab)`
`=(a+b+c)(a^2+b^2+c^2−ab−bc−ca)=VP`
a: \(\left(a+b\right)\left(a^2-b^2\right)+\left(b-c\right)\left(b^2-c^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)
\(=a^3-ab^2+a^2b-b^3+b^3-bc^2-b^2c+c^3+\left(c+a\right)\left(c^2-a^2\right)\)
\(=a^3+c^3-ab^2-b^2c+a^2b-bc^2+\left(c+a\right)\left(c+a\right)\left(c-a\right)\)
\(=\left(c+a\right)\left(c^2-ac+a^2\right)-b^2\left(c+a\right)-b\left(c-a\right)\left(c+a\right)+\left(c+a\right)^2\cdot\left(c-a\right)\)
=(c+a)\(\left(c^2-ac+a^2-b^2-bc+ba+c^2-a^2\right)\)
=(c+a)\(\left(2c^2-2a^2-b^2-ac-bc+ba\right)\)
b: \(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(=a^3\left(b-c\right)+b^3\left(c-b+b-a\right)+c^3\left(a-b\right)\)
\(=a^3\left(b-c\right)-b^3\left(b-c\right)-b^3\left(a-b\right)+c^3\left(a-b\right)\)
\(=\left(b-c\right)\left(a^3-b^3\right)-\left(a-b\right)\left(b^3-c^3\right)\)
=(b-c)(a-b)\(\left(a^2+ab+b^2-b^2+bc-c^2\right)\)
=(b-c)(a-b)\(\left(a^2+ab+bc-c^2\right)\)
=(b-c)(a-b)\(\left\lbrack\left(a-c\right)\left(a+c\right)+b\left(a+c\right)\right\rbrack\)
=(b-c)(a-b)(a+c)(a-c+b)
a: Sửa đề: A=ab(a-b)+bc(b-c)+ca(c-a)
\(=a^2b-ab^2+b^2c-bc^2+ca\left(c-a\right)\)
\(=b\left(a^2-c^2\right)-b^2\left(a-c\right)-ac\left(a-c\right)\)
=b(a-c)(a+c)\(-b^2\left(a-c\right)-ac\left(a-c\right)\)
=(a-c)\(\left(ba+bc-b^2-ac\right)\)
=(a-c)[b(a-b)-c(a-b)]
=(a-c)(a-b)(b-c)
c: \(C=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b+c-a\right)\left\lbrack\left(a+b+c\right)^2+a\left(a+b+c\right)+a^2\right\rbrack-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(a^2+b^2+c^2+2ab+2ac+2bc+a^2+ab+ac+a^2\right)\) -(b+c)(\(b^2-bc+c^2\) )
=(b+c)\(\left(3a^2+b^2+c^2+3ab+3ac+2bc-b^2+bc-c^2\right)\)
=(b+c)(\(3a^2+3ab+3ac+3bc\) )
=3(b+c)[a(a+b)+c(a+b)]
=3(b+c)(a+b)(a+c)
d) Ta có: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)\cdot c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)