Dec 30, 2013 Thread starter #1 Albert Well-known member Jan 25, 2013 1,225 $a+b+c+d=0$ $a^3+b^3+c^3+d^3=5$ $find: abc+abd+acd+bcd=?$
Dec 30, 2013 Admin #2 anemone MHB POTW Director Staff member Feb 14, 2012 3,967 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 $\begin{align*} abc+abd+acd+bcd&=-(ab(a+b)+cd(c+d))\\&=-(-\dfrac{5}{3})=\dfrac{5}{3} \end{align*}$
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 $\begin{align*} abc+abd+acd+bcd&=-(ab(a+b)+cd(c+d))\\&=-(-\dfrac{5}{3})=\dfrac{5}{3} \end{align*}$
Dec 31, 2013 Thread starter #3 Albert Well-known member Jan 25, 2013 1,225 Re: find: abc+abd+acd+bcd=? my solution : Spoiler let $a+b=x---(1),\,\,c+d=-x---(2)$ $(1)^3+(2)^3=a^3+b^3+c^3+d^3+3a^2b+3ab^2+3c^2d+3cd^2=0$ $5+3ab(a+b)+3cd(c+d)=5+3abx-3cdx=0$ $5=3(cdx-abx)---(3)$ $abc+abd+acd+bcd=ab(c+d)+cd(a+b)=cdx-abx=\dfrac{5}{3}---from(3)$
Re: find: abc+abd+acd+bcd=? my solution : Spoiler let $a+b=x---(1),\,\,c+d=-x---(2)$ $(1)^3+(2)^3=a^3+b^3+c^3+d^3+3a^2b+3ab^2+3c^2d+3cd^2=0$ $5+3ab(a+b)+3cd(c+d)=5+3abx-3cdx=0$ $5=3(cdx-abx)---(3)$ $abc+abd+acd+bcd=ab(c+d)+cd(a+b)=cdx-abx=\dfrac{5}{3}---from(3)$