Proof problem

If $$a+b+c+d=0$$,prove that: $abc+bcd+cda+dab=\sqrt {(bc-ad)(ca-bd)(ab-cd)}.$

Note by Rohit Udaiwal
2 years, 8 months ago

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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.

- 2 years, 8 months ago

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Nice use of that identity!!

- 2 years, 8 months ago

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