Question

Chứng minh rằng:

a) Nếu a+b+c=0 thì \(a^3+b^{^3}+c^3\)=3abc

b)Nếu a+b+c+d=0 thì \(a^3+b^3+c^3+d^3=3\left(c+d\right)\left(ab-cd\right)\)

Answer 1

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

⇔ \(a^3+b^3+c^3-3abc=0\)

⇔ \(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

⇔ \(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)

⇔\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

⇔ \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

Với : a + b + c = 0 thì dễ thấy đẳng thức trên đúng .

Từ đó suy ra : đpcm .

\(b.a+b+c+d=0\)

⇔ \(a+b=-\left(c+d\right)\)

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

⇔ \(a^3+b^3+3a^2b+3ab^2=-\left(c^3+3c^2d+3cd^2+d^3\right)\)

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

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

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

⇔ \(a^3+b^3+c^3+d^3=3\left(c+d\right)\left(ab-cd\right)\) ( đpcm)