write as
\[a=-(b+c+d)\]
\[a(bc+cd+db)+bcd=-((b+c+d)(bc+cd+db)-bcd)\]
by daniel lui's identity this is
\[-(b+c)(c+d)(d+b)\]
this is
\[-(-a-d)(-a-b)(-a-c)=(a+b)(a+c)(a+d)\]
square
\[((a+b)(a+d))((a+d)(a+c))((a+c)(a+b))=(a^2+(b+d)a+bd)(a^2+(d+c)a+dc)(a^2+(c+b)a+cb)\]
\[=(a^2+(-a-c)a+bd)(a^2+(-a-b)a+dc)(a^2+(-a-d)a+cb)=(bd-ac)(dc-ab)(bc-ad)\]
so
\[abc+bcd+cda+dab=\pm\sqrt{(bd-ac)(dc-ab)(bc-ad)}\]
note ther should pe \(\pm\) as the LHS can be negative.
–
Aareyan Manzoor
·
1 year, 4 months ago

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TopNewestwrite as \[a=-(b+c+d)\] \[a(bc+cd+db)+bcd=-((b+c+d)(bc+cd+db)-bcd)\] by daniel lui's identity this is \[-(b+c)(c+d)(d+b)\] this is \[-(-a-d)(-a-b)(-a-c)=(a+b)(a+c)(a+d)\] square \[((a+b)(a+d))((a+d)(a+c))((a+c)(a+b))=(a^2+(b+d)a+bd)(a^2+(d+c)a+dc)(a^2+(c+b)a+cb)\] \[=(a^2+(-a-c)a+bd)(a^2+(-a-b)a+dc)(a^2+(-a-d)a+cb)=(bd-ac)(dc-ab)(bc-ad)\] so \[abc+bcd+cda+dab=\pm\sqrt{(bd-ac)(dc-ab)(bc-ad)}\] note ther should pe \(\pm\) as the LHS can be negative. – Aareyan Manzoor · 1 year, 4 months ago

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– Rohit Udaiwal · 1 year, 4 months ago

Nice use of that identity!!Log in to reply