\(\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3\)

\(=\left(a+b\right)^3+3\cdot c\cdot\left(a+b\right)^2+3\cdot c^2\left(a+b\right)+c^3\)

\(=a^3+3a^2b+3ab^2+b^3+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2+c^3\)

\(=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

thay a^3+b^3=(a+b)^3 -3ab(a+b) .Ta có : 

a^3+b^3+c^3-3abc=0 

<=>(a+b)^3 -3ab(a+b) +c^3 - 3abc=0 

câu 2:<=>[(a+b)^3 +c^3] -3ab.(a+b+c)=0 

<=>(a+b+c). [(a+b)^2 -c.(a+b)+c^2] -3ab(a+b+c)=0 

<=>(a+b+c).(a^2+2ab+b^2-ca-cb+c^2-3ab)... 

<=>(a+b+c).(a^2+b^2+c^2-ab-bc-ca)=0 

luôn đúng do a+b+c=0

câu 1:(a+b+c)^3=((a+b)+c)^3=(a+b)^3+c^3+3(a+b)c(a+b+c)
=a^3+b^3+3ab(a+b)+c^3+3(a+b)c(a+b+c)
=a^3+b^3+c^3+3(a+b)(ab+c(a+b+c))
=a^3+b^3+c3^+3(a+b)(ab+ac+bc+c2)
=a^3+b^3+c^3+3(a+b)(a+c)(b+c)

CHÚC BẠN HỌC TỐT^^

mk viết viết đề nha 

=a³+3a²b +3ab²+b³+3(a+b)².c+3.(a+b).c²+c³

= a³+b³+c³+[ 3a²b+3a²b+3(a+b)³.c+3.(a+b).c^2]

= a³+b³+c³+[3ab(a+b)+3(a+b)²c+3(a+b)c²]

= a³+b³+c³+3(a+b)[ ab+(a+b)c+c²]

= a³+b³+c³+3(a+b)(ab+ac+bc+c²)

= a³+b³+c³+3(a+b)[a(b+c)+c.(b+c)]

= a³+b³+c³+3(a+b)(b+c)(a+c)

=> dpcm 

ta có (a+b)^3 =a^3 +b^3 +3ab(a+b) 

=>[(a+b) +c ]^3 =(a+b)^3 +c^3 +3c(a+b)[a+b+c) 
[(a+b) +c ]^3 = a^3+b^3 +3ab(a+b) +3c(a+b)(a+b+c)+c^3 
[(a+b) +c ]^3 =a^3+b^3+c^3 +3(a+b)[ab+c.(a+b+c) ] 
[(a+b) +c ]^3 = a^3+b^3+c^3 +3(a+b)[ ab+ca+cb+c^2] 
[(a+b) +c ]^3 = a^3+b^3+c^3 +3(a+b)[ a(c+b) +c(b+c)] 
[(a+b) +c ]^3 =a^3+b^3+c^3 +3(a+b)(b+c)(a+c) (vế trái)

Điều cần chứng minh giờ thì đã sáng tỏ! ^_^

(a+b+c)³=(a+b)³+3(a+b)²c+3(a+b)c²+c³

=a³+b³+3ab.(a+b)+3(a+b)²c+3(a+b)c²+c³

=a³+b³+c³+3(a+b)(ab+ac+bc+c²)

=a³+b³+c³+3(a+b)[a.(b+c)+c.(b+c)]

=a³+b³+c³+3(a+b)(b+c)(c+a) 

=>dpcm

P=12(5^2+1)(5^4+1)(5^8+1)(5^16+1)

=>2P=24(5^2+1)(5^4+1)(5^8+1)(5^16+1)

=(5²-1)(5²+1)(5⁴+1)(5⁸+1)(5¹⁶+1)

=(5⁴-1)(5⁴+1)(5⁸+1)(5¹⁶+1)

=(5⁸-1)(5⁸+1)(5¹⁶+1)

=(5¹⁶-1)(5¹⁶+1)

=5³²-1

==>P=(5³²-1)/2

thanks anh Triều!

Theo bất đẳng thức Cauchy-Schwarzt ta có \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ca}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}.\)
Mặt khác, \(a^2+b^2+c^2\ge ab+bc+ca\), do đó ta suy ra \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2.\)

P=\(\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ca}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2}=a^2+b^2+c^2\)

\(\frac{a^3}{b}+\frac{a^3}{b}+b^2\ge3\sqrt[3]{\frac{a^3}{b}.\frac{a^3}{b}.b^2}=3a^2\)