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a+b+c+d=0
=> a + b = -(c+d)
=> (a+b)^3 = -(c+d)^3
=> a^3 + b^3 + 3ab (a+b) = -c^3- d^3 - 3cd (c+d)
=> a^3+b^3+c^3+d^3 = -3ab (a+b) - 3cd (c+d)
=> a^3 + b^3 + c^3 + d^3 = 3ab (c+d)- 3cd (c+d) [vì a+b = - (c+d)]
==> a^3 + b^^3 + c^3 + d^3 =3 (c+d) (ab-cd) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
a+b+c+d=0
=>a+b = - (c+d)
=> (a+b)^3= - (c+d)^3
=> a^3 + b^3 + 3ab(a+b) = - c^3 - d^3 - 3cd(c+d)
=> a^3 + b^3 + c^3 + d^3 = - 3ab(a+b) - 3cd(c+d)
=> a^3 + b^3 + c^3 + d^3 = 3ab(c+d) - 3cd(c+d) ( Vì a+b = - (c+d))
==> a^3 + b^3 + c^3 + d^3 = 3(c+d)(ab-cd) (đpcm).
![](https://rs.olm.vn/images/avt/0.png?1311)
a+b+c+d=0
=>a+d=-(b+c)
=>(a+d)^3=-(b+c)^3
=>\(a^3+d^3+3ad\left(a+d\right)=-b^3-c^3-3bc\left(b+c\right)\)
=>\(a^3+d^3+3ad\left(a+d\right)=-b^3-c^3+3bc\left(a+d\right)\)
=>\(a^3+d^3+b^3+c^3=3bc\left(a+d\right)-3ad\left(a+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(a+d\right)\left(bc-ad\right)\)
=>\(a^3+b^3+c^3+d^3=3\left(b+c\right)\left(ad-bc\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: a+b+c+d=0
\(\Leftrightarrow b+c=-\left(a+d\right)\)
\(\Leftrightarrow\left(b+c\right)^3=-\left(a+d\right)^3\)
\(\Leftrightarrow b^3+c^3+3bc\left(b+c\right)=-\left[a^3+d^3+3ad\left(a+d\right)\right]\)
\(\Leftrightarrow b^3+c^3+3bc\left(b+c\right)=-a^3-d^3-3ad\left(a+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3bc\left(b+c\right)-3ad\left(a+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3bc\left(b+c\right)-3ad\cdot\left[-\left(b+c\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3bc\left(b+c\right)+3ad\left(b+c\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(b+c\right)\left(ad-bc\right)\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: a+b+c+d=0
⇔\(a+d=-\left(b+c\right)\)
\(\Leftrightarrow\left(a+d\right)^3=-\left(b+c\right)^3\)
\(\Leftrightarrow a^3+d^3+3ad\left(a+d\right)=-\left[b^3+c^3+3bc\left(b+c\right)\right]\)
\(\Leftrightarrow a^3+d^3+3ad\left(a+d\right)=-b^3-c^3-3bc\left(b+c\right)\)
\(\Leftrightarrow a^3+d^3+b^3+c^3=-3ad\left(a+d\right)-3bc\left(b+c\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ad\left(a+d\right)+3bc\left(a+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=\left(a+d\right)\left(-3ad+3bc\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=\left(a+d\right)\cdot3\cdot\left(-ad+bc\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-\left(b+c\right)\cdot3\cdot\left[-\left(ad-bc\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\cdot\left(b+c\right)\cdot\left(ad-bc\right)\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có : a+b+c+d=0
=>a+b=-(c+d)
=> (a+b)3=-(c+d)3
=> a3+b3+3ab(a+b)=-c3-d3-3cd(c+d)
=> a3+b3+c3+d3=-3ab(a+b)-3cd(c+d)
=> a3+b3+c3+d3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d))
=> a3 +b3+c3+d3==3(c+d)(ab-cd)
(dpcm)
ta có a+b+c+d = 0=> b+c= -( a+d) => (b+c)^3 = - (a+d)^3
=> b^3+ c^3 + 3bc( b+c) = -( a^3 +d^3 + 3ad(a+d))
=> a^3+b^3+c^3+d^3 = - 3ad( a+d) - 3bc(b+c) = 3ad(b+c) - 3bc(b+c)
= 3(b+c)(ad-bc)
sao cậu tự đặt câu hỏi rồi lại tự trả lời luôn
thế là sao??????????