Proof problem

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

#Algebra

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 5 years, 6 months ago

Nice use of that identity!!

Rohit Udaiwal - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$