Find: abc + abd + acd + bcd = ?

[Albert1]

$a+b+c+d=0$

$a^3+b^3+c^3+d^3=5$

$find: abc+abd+acd+bcd=?$

 

[anemone]

Re: find: abc+abd+acd+bcd=?

My solution:

Spoiler

$\begin{align*} (a+b+c+d)^3&=((a+b)+(c+d))^3\\&=(a+b)^3+3(a+b)(c+d)(a+b+c+d)+(c+d)^3\\&= a^3+3ab(a+b)+b^3+3(a+b)(c+d)(0)+c^3+3cd(c+d)+d^3\\&=a^3+b^3+c^3+d^3+3(ab(a+b)+cd(c+d)) \end{align*}$

$\therefore 0^3=5+3(ab(a+b)+cd(c+d))\;\;\;\rightarrow ab(a+b)+cd(c+d)=-\dfrac{5}{3}$

Notice that

$\begin{align*} abc+abd+acd+bcd&=ab(c+d)+cd(a+b)\\&=ab(-a-b)+cd(-c-d)\\&=-(ab(a+b)+cd(c+d)) \end{align*}$

Therefore we get

$\begin{align*} abc+abd+acd+bcd&=-(ab(a+b)+cd(c+d))\\&=-(-\dfrac{5}{3})=\dfrac{5}{3} \end{align*}$

 

[Albert1]

Re: find: abc+abd+acd+bcd=?

my solution :

Spoiler

let $a+b=x---(1),\,\,c+d=-x---(2)$
$(1)^3+(2)^3=a^3+b^3+c^3+d^3+3a^2b+3ab^2+3c^2d+3cd^2=0$
$5+3ab(a+b)+3cd(c+d)=5+3abx-3cdx=0$
$5=3(cdx-abx)---(3)$
$abc+abd+acd+bcd=ab(c+d)+cd(a+b)=cdx-abx=\dfrac{5}{3}---from(3)$