Question

Chứng minh các bất đẳng thức :

Cho a + b + c = 0 . Chứng minh rằng :  a³ + b³ + c³ = 3abc

Answer 1

ta có:

(a+b+c)³=a³+b³+c³+3a²b+3a²c+3b²a+3b²c+3c²a+3c²b+6abc

=a³+b³+c³+(3a²b+3a²c+3abc)+(3b²a+3b²c+3abc)+(3c²a+3c²b+3abc)-3abc

=a³+b³+c³+3a.(ab+ac+bc)+3b(ab+ac+bc)+3c.(ab+ac+bc)-3abc

=a³+b³+c³+3.(a+b+c)(ab+ac+bc)-3abc

=>0³=a³+b³+c³+3.0.(ab+ac+bc)-3abc

0=a³+b³+c³-3abc

<=>a³ + b³ + c³ = 3abc

Answer 2

a + b + c = 0 =>  a + b  = -c 

TA có 

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

Thay  a +b = -c ta có 

a^3 + b^3 + c^3 = -c^3 - 3ab.(-c) + c^3 = 3abc (ĐPCM)

Answer 3

a+b+c=0

=>(a+b+c)³=0

=>a³+b³+c³+3a²b+3ab²+3b²c+3bc²+3a²c+3ac²+6abc=0

=>a³+b³+c³+(3a²b+3ab²+3abc)+(3b²c+3bc²+3abc)+(3a²c+3ac²+3abc)-3abc=0

=>a³+b³+c³+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)=3abc

Do a+b+c=0

=>a³+b³+c³=3abc(ĐPCM)