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c: Ta có: \(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
\(=a^4-2a^3b+2ab^3-b^4\)
\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)-2ab\left(a^2-b^2\right)\)
\(=\left(a-b\right)^3\cdot\left(a+b\right)\)
b) Ta có: \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2\)
\(=\left(ab^2-cb^2\right)+\left(ca^2-c^2a\right)+\left(bc^2-ba^2\right)\)
\(=b^2\left(a-c\right)+ca\left(a-c\right)+b\left(c^2-a^2\right)\)
\(=\left(a-c\right)\left(b^2+ca\right)-b\left(a-c\right)\left(a+c\right)\)
\(=\left(a-c\right)\left(b^2+ca-ba-bc\right)\)
\(=\left(a-c\right)\left[b\left(b-a\right)+c\left(a-b\right)\right]\)
\(=\left(a-c\right)\left[b\left(b-a\right)-c\left(b-a\right)\right]\)
\(=\left(a-c\right)\left(b-a\right)\left(b-c\right)\)
a)
$A=\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}$
$\ge \dfrac{(a+b+c)^2}{a(b^2+1)+b(c^2+1)+c(a^2+1)}\qquad (\text{Cauchy Engel})$
$=\dfrac{1}{ab^2+bc^2+ca^2+1}$
$\ge \dfrac{1}{ab(a+b)+bc(b+c)+ca(c+a)+1}$
$=\dfrac{1}{(a+b+c)(ab+bc+ca)+1}
$\ge \dfrac{1}{\frac13+1}$ $=\dfrac34$
Dấu bằng khi $a=b=c=\dfrac13$.
$\boxed{A_{\min}=\dfrac34}$
b)
$B=\dfrac{a}{ab+2c}+\dfrac{b}{bc+2a}+\dfrac{c}{ca+2b}$
$\ge \dfrac{(a+b+c)^2}{a(ab+2c)+b(bc+2a)+c(ca+2b)}$
$=\dfrac4{a^2b+b^2c+c^2a+2(ab+bc+ca)}$
Lại có $a^2b+b^2c+c^2a\le (a+b+c)(ab+bc+ca)$$=2(ab+bc+ca)$
Nên $B\ge \dfrac4{4(ab+bc+ca)}$$=\dfrac1{ab+bc+ca}$
$\ge \dfrac1{\frac{(a+b+c)^2}{3}}$ $=\dfrac34$
Dấu bằng khi $a=b=c=\dfrac23$.
$B_{\min}=\dfrac34$
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)
Ta có x + y = a + b
=> (x + y)2 = (a + b)2
=> x2 + y2 + 2xy = a2 + b2 + 2ab
=> xy = ab
Lại có x + y = a + b
=> (x + y)3 = (a + b)3
=> x3 + 3x2y + 3xy2 + y3 = a3 + 3a2b + 3ab2 + b3
=> x3 + y3 + 3xy(x + y) = a3 + b3 + 3ab(a + b)
=> x3 + y3 = a3 + b3 (vì x + y = a + b ; xy = ab)