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a+b+c+d=0

=>a+d=-(b+c)

=>(a+d)^3=-(b+c)^3

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

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

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

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

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

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)

27 tháng 6 2016

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)

13 tháng 7 2016

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 )

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

25 tháng 8 2021

a+b+c+d=0 => a+d= -b-c;       (a+b)3=a3+b3+3ab(a+b) => a3+b3=(a+b)3-3ab(a+b)

a3+d3+b3+d3

=(a+d)3- 3ad(a+d)+ (b+c)3-3bc(b+c) (1)

Do a+d=-b-c nên pt (1) trở thành:

-(b+c)3-3ad(-b-c)+ (b+c)3-3bc(b+c)

=3ad(b+c)-3bc(b+c)

=3(b+c)(ad-bc) <đccm>

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)

DD
1 tháng 8 2021

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

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

\(=-3bc\left(b+c\right)+3ad\left(b+c\right)\)

\(=3\left(b+c\right)\left(ad-bc\right)\)