Question

Chứng minh :
a) ( a + b + c )³ - a³ - b³ -c³ = 3( a + b )( b +c )( c + a )
b) a³ +b³ + c³ = 3abc nếu a + b + c = 0

Answer 1

Lời giải:

a)

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

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

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

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

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

\(=3(a+b)(ab+ca+bc+c^2)=3(a+b)[a(b+c)+c(b+c)]\)

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

b)

Áp dụng kết quả phần a: Nếu $a+b+c=0$ thì:

\(0^3-a^3-b^3-c^3=3(0-c)(0-a)(0-b)\)

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

\(\Leftrightarrow a^3+b^3+c^3=3abc\)