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

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

Đặt \(P=\dfrac{a^3}{a^2+b^2+ab}+\dfrac{b^3}{b^2+c^2+bc}+\dfrac{c^3}{c^2+a^2+ca}\)

Ta có: \(\dfrac{a^3}{a^2+b^2+ab}=a-\dfrac{ab\left(a+b\right)}{a^2+b^2+ab}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^3b^3}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)

Tương tự: \(\dfrac{b^3}{b^2+c^2+bc}\ge\dfrac{2b-c}{3}\) ; \(\dfrac{c^3}{c^2+a^2+ca}\ge\dfrac{2c-a}{3}\)

Cộng vế:

\(P\ge\dfrac{a+b+c}{3}=673\)

Dấu "=" xảy ra khi \(a=b=c=673\)

a^3+b^3+c^3-3abc

=(a+b)^3+c^3-3ab(a+b)-3bca

=(a+b+c)(a^2+2ab+b^2-ac-bc+c^2)-3ab(a+b+c)

=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)

\(\left(a+b+c\right)^2=a^2+b^2+c^2\)

=>\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)

=>\(2\left(ab+bc+ac\right)=0\)

=>ab+bc+ac=0

\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)

=>\(\dfrac{\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3}{\left(abc\right)^3}=\dfrac{3}{abc}\)

=>\(\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3=3\left(abc\right)^2\)

\(\Leftrightarrow\left(ab+bc\right)^3-3\cdot ab\cdot bc\cdot\left(ab+bc\right)+\left(ac\right)^3=3\left(abc\right)^2\)

=>\(\left(-ac\right)^3-3\cdot ab\cdot bc\cdot\left(-ac\right)+\left(ac\right)^3-3\left(abc\right)^2=0\)

=>\(-a^3c^3+a^3c^3+3a^2b^2c^2-3a^2b^2c^2=0\)

=>0=0(đúng)

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)

ké ý (b) ạ!!!

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)

1, C/m : a^3 + b^3 + c^3 ≥ a^2.căn (bc) + b^2.căn (ac) + c^2.căn (ab) 
Ta có : 2( a^3 + b^3 + c^3 ) = ( a^3 + b^3 + c^3 ) + ( a^3 + b^3 + c^3 ) 
≥ 3abc + a^3 + b^3 + c^3 ( BĐT Côsi ) 
= a^3 + abc + b^3 + abc + c^3 + abc ≥ 2.a^2.căn (bc) + 2.b^2.căn (ac) + 2.c^2.căn (ab) ( BĐT Côsi ) 
=> a^3 + b^3 + c^3 ≥ a^2.căn (bc) + b^2.căn (ac) + c^2.căn (ab) 
Dấu " = " xảy ra khi a = b = c. 


2, C/m : (a^2 + b^2 + c^2)(1/(a + b ) + 1/(b + c) +1/(a + c) ) ≥ (3/2)(a + b + c) ( 1 ) 
Áp dụng BĐT Bunhiacốpxki cho phân số ( :D ) ta được : 
(a^2 + b^2 + c^2)(1/(a + b ) + 1/(b + c) +1/(a + c) ) ≥ (a^2 + b^2 + c^2).[(1+1+1)^2/(a+b+b+c+a+c)] = (a^2 + b^2 + c^2) . 9/[2.(a+b+c)] 
(1) <=> (a^2 + b^2 + c^2) . 9/[2.(a+b+c)] ≥ (3/2)(a + b + c) 
<=> 3(a^2 + b^2 + c^2) ≥ (a + b + c)^2 
<=> a^2 + b^2 + c^2 ≥ ab + bc + ca. 
BĐT cuối đúng nên => đpcm ! 
Dấu " = " xảy ra khi a = b = c. 


3, C/m : a^4 + b^4 + c^4 ≥ (a + b + c)abc 
Ta có : 2( a^4 + b^4 + c^4 ) = (a^4 + b^4 +c^4) + (a^4 + b^4 +c^4) 
≥ ( a^2.b^2 + b^2.c^2 + c^2.a^2 ) + (a^4 + b^4 +c^4) = ( a^4 + b^2.c^2 ) + ( b^4 + c^2.a^2 ) + ( c^4 + a^2.b^2 ) 
≥ 2.a^2.bc + 2.b^2.ca + 2.c^2.ab ( BĐT Côsi ) 
= 2.abc(a + b + c) 
Do đó a^4 + b^4 + c^4 ≥ (a + b + c)abc 
Dấu " = " xảy ra khi a = b = c.