a ) \(a^3+b^3+c^3=3abc\)

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

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

Nếu : \(a+b+c=0\) thì đẳng thức trên đúng .

\(\Rightarrowđpcm\)

b ) \(a+b+c+d=0\)

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

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

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

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

Chúc bạn học tốt !!!

a ) a^3+b^3+c^3=3abca3+b3+c3=3abc

\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0⇔(a+b)3+c3−3ab(a+b)−3abc=0

\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0⇔(a+b+c)(a2+b2+c2−ab−bc−ac)=0

Nếu : a+b+c=0a+b+c=0 thì đẳng thức trên đúng .(đpcm)

b ) a+b+c+d=0a+b+c+d=0

\Rightarrow a+b=-\left(c+d\right)\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3⇒a+b=−(c+d)⇔(a+b)3=−(c+d)3

\Leftrightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)⇔a3+b3+c3+d3=−3ab(a+b)−3cd(c+d)

\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)⇔a3+b3+c3+d3=3ab(c+d)−3cd(c+d)

\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(cb-cd\right)\left(đpcm\right)⇔a3+b3+c3+d3=3(c+d)(cb−cd)(đpcm)

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

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

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

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

Nếu : \(a+b+c=0\) thì đẳng thức trên đúng .

\(\Rightarrowđpcm\)

\(b.a+b+c+d=0\Rightarrow a+b=-\left(c+d\right)\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)

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

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

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

\(a+b+c+d=0\Rightarrow a+b=-\left(c+d\right)\)

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

\(\Rightarrow\left(a+b\right)^3+\left(c+d\right)^3=0\)

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

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

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

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

Ta có : \(a+b+c+d=0\Leftrightarrow a+d=-\left(b+c\right)\)

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

\(\Leftrightarrow a^3+d^3+3ad\left(a+d\right)=-\left[c^3+b^3+3bc\left(b+c\right)\right]\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ad\left(a+d\right)-3bc\left(b+c\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3ad\left(b+c\right)-3bc\left(b+c\right)\) (vì a + d = - b - c )

\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(b+c\right)\left(ad-bc\right)\)

Theo đề, a+b+c+d=0

\(\Rightarrow a+b=-\left(c+d\right)\)

Ta có: \(VT=\left(a+b\right)\left(a^2-ab+b^2\right)+\left(c+d\right)\left(c^2-cd+d^2\right)\)

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

Để có ĐPCM ta xét hiệu: \(c^2-cd+d^2-a^2+ab-b^2-3\left(ab+cd\right)=c^2-4cd+d^2-a^2-2ab-b^2=c^2-4cd+d^2-\left(a+b\right)^2=c^2-4cd+d^2-\left(c+d\right)^2=-6cd\)

S nó ko = 0 ta:::xem lại đề..Hay mk lm sai j đó

a +b+c+d=0

0 nha bn

Ta có : \(a+b+c+d\)\(=0\)

\(\Rightarrow a+b=-c-d\)hoặc \(a+c=-b-d\)

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

\(\Rightarrow a^3+c^3+3ac\left(a+c\right)=-b^3-d^3-3bd\left(b+d\right)\)

\(\Rightarrow a^3+b^3+c^3+d^3=-3ac\left(a+c\right)-3bd\left(b+d\right)\)

                                            \(=-3ac\left(a+c\right)-3ac\left(b+d\right)+3ac\left(b+d\right)-3bd\left(b+d\right)\)

                                             \(=-3ac\left(a+b+c+d\right)+3\left(b+d\right)\left(ac+bd\right)\)

                                               \(=-3ac.0+3\left(ac-bd\right)\left(b+d\right)\)

                                               \(=3\left(ac-bd\right)\left(b+d\right)\)

Vậy ta đã chứng minh được : \(a^3+b^3+c^3+d^3=3\left(ac-bd\right)\left(b+d\right)\)

a) a² + b² + c2 = ab + ac + bc

=> 2a² + 2b² + 2c² = 2ab + 2ac + 2bc

=> 2a² + 2b² + 2c² - 2ab - 2ac - 2bc = 0

=> (a² - 2ab + b²) + (a² - 2ac + c²) + (b² - 2bc + c²) = 0

=> (a - b)² + (a - c)² + (b - c)² = 0 

Do 3 hạng tử trên đều có giá trị lớn hơn hoặc bằng 0 nên a - b = a - c = b - c = 0

=> a = b = c 

b) a³ + b³ + c³ = 3abc

=> a³ + b³ + c³ - 3abc = 0

=> a³ + 3a²b + 3ab² + b³ + c³ - 3abc - 3a²b - 3ab² = 0

=> (a + b)³ + c³ - 3ab(a + b + c) = 0

=> (a + b + c)(a² + 2ab + b² - bc - ac + c²) - 3ab(a + b + c) = 0

=> (a + b + c)(a² + b² + c² - ab - bc - ac) = 0 

=> a + b + c = 0

hoặc a² + b² + c² = ab + bc + ac =>  a = b = c

a)\(a^2+b^2+c^2=ab+bc+ca\)\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Rightarrow a=b=c}\)

b)\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)

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

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

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

\(\Rightarrow\hept{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)