Để chứng minh hằng đẳng thức a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a) = (a+b+c)^3, ta sẽ sử dụng công thức khai triển đa thức.

Theo công thức khai triển đa thức, ta có:

(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a)

Vậy, hằng đẳng thức được chứng minh.

bạn vào link mik gửi nha!!!

(a+b+c)³ = (a + b)³ + c³ + 3(a+b)c.(a+b+c)

= a³ + b³ + 3ab.(a+b) + c³ + 3(a+b)c(a+b+c) = 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).(a+c).(b+c)\(\Rightarrowđpcm\)

(Cái trong ngoặc bạn tự phân tích đa thức thành nhân tử 1 cách dễ dàng.
(a+b+c)3−a

Ta có: (a+b+c)³ - a³ -b³ -c³

=a³ +b³ +c³ +3(a+b)(b+c)(c+a)-a³-b³-c³

=(a³-a³)+(b³-b³)+(c³-c³) +3(a+b)(b+c)(c+a)

=3(a+b)(b+c)(c+a) (đpcm)

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

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

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

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

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

b)  \(VT=a^3+b^3+c^3-3abc\)

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

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

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

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\)

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

= a³ + b³ + c³  + 3abc + 3ac² + 3b²c + 3bc² + 3a²b + 3a²c + 3b²a + 3abc

= ( a + b )³ + 3( a+b)²c + 3(a+b)c² + c³

= (a+b+c)³

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

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

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

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

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

\(=a^3+b^3+c^3+3\left(a+b\right)\left[ab+\left(a+b\right)c+c^2\right]\)

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

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

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

(a+b+c)³=a³+b³+c³+3(a+b)(b+c)(c+a)

<=>(a+b+c)³-a³-b³-c³=3(a+b)(b+c)(c+a) (1)

Ta có:(a+b+c)³-a³-b³-c³=[(a+b+c)³-a³]-(b³+c³)

=(a+b+c-a)(a²+b²+c²+2ab+2bc+2ca+a²+ab+ac+a²)-(b+c)(b²-bc+c²)

=(b+c)(3a²+b²+c²+3ab+3ac+2bc)-(b+c)(b²-bc+c²)

=(b+c)(3a²+b²+c²+3ab+3ac+2bc-b²+bc-c²)

=(b+c)(3a²+3ab+3ac+3bc)

=3(b+c)](a²+ab)+(ac+bc)]

=3(b+c)[a(a+b)+c(a+b)]

=3(b+c)(a+c)(a+b)

=>(1) đúng => đpcm

biến đổi vế trái :

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

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

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

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

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

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

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

Vế trái bằng vế phải đẳng thức được chứng minh.

\(a^3+b^3+c^3-3abc\)

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

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

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

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(=\frac{1}{2}\left(a+b+c\right).2\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(=\frac{1}{2}\left(a+b+c\right)[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)]\)

\(=\frac{1}{2}\left(a+b+c\right)[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2]\)

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{15}+1\right)\)

\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(\frac{1}{2}\left(5^{32}+1\right)=\frac{5^{32}+1}{2}\)

a)

 Ta có

a chia 5 dư 4

=> a=5k+4 ( k là số tự nhiên )

\(\Rightarrow a^2=\left(5k+4\right)^2=25k^2+40k+16\)

Vì 25k^2 chia hết cho 5

    40k chia hết cho 5

    16 chia 5 dư 1

=> đpcm

2) Ta có

\(12=\frac{5^2-1}{2}\)

Thay vào biểu thức ta có

\(P=\frac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)}{2}\)

\(\Rightarrow P=\frac{\left[\left(5^2\right)^2-1^2\right]\left[\left(5^2\right)^2+1^2\right]\left(5^8+1\right)}{2}\)

\(\Rightarrow P=\frac{\left[\left(5^4\right)^2-1^2\right]\left[\left(5^4\right)^2+1^2\right]}{2}\)

\(\Rightarrow P=\frac{5^{16}-1}{2}\)

3)

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

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

\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+cb+c^2\right)\)

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