If x + y + z = xyz, the ∑3x−x31−3x2=

Let x=tan⁡A,y=tan⁡B,z=tan⁡C. Then x+y+z=xyz ⇒A+B+C=π⇒ 3A+3B=3π−3C⇒ tan⁡(3A+3B)=tan⁡(3π−3C)⇒ tan⁡3A+tan⁡3B1−tan⁡3Atan⁡3B=−tan⁡3C⇒ tan⁡3A+tan⁡3B+tan⁡3C=tan⁡3Atan⁡3Btan⁡3C⇒ ∑3tan⁡A−tan3⁡A1−3tan2⁡A=∏3tan⁡A−tan3⁡A1−3tan2⁡A⇒ ∑3x−x31−3x2=∏3x−x31−3x2

If x + y + z = xyz, the ∑3x−x31−3x2=

Let x=tan⁡A,y=tan⁡B,z=tan⁡C. Then x+y+z=xyz ⇒A+B+C=π⇒ 3A+3B=3π−3C⇒ tan⁡(3A+3B)=tan⁡(3π−3C)⇒ tan⁡3A+tan⁡3B1−tan⁡3Atan⁡3B=−tan⁡3C⇒ tan⁡3A+tan⁡3B+tan⁡3C=tan⁡3Atan⁡3Btan⁡3C⇒ ∑3tan⁡A−tan3⁡A1−3tan2⁡A=∏3tan⁡A−tan3⁡A1−3tan2⁡A⇒ ∑3x−x31−3x2=∏3x−x31−3x2

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If x + y + z = xyz, the $\sum (\frac{3 x - x^{3}}{1 - 3 x^{2}}) =$

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3xyz

$\prod (\frac{3 x - x^{3}}{1 - 3 x^{2}})$

$\frac{1}{(3 x y z)}$

answer is A.

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Detailed Solution

$\begin{array}{l} Let x = \tan A, y = \tan B, z = \tan C . Then \\ x + y + z = xyz \Rightarrow A + B + C = π \\ \Rightarrow 3 A + 3 B = 3 π - 3 C \\ \Rightarrow \tan (3 A + 3 B) = \tan (3 π - 3 C) \\ \Rightarrow \frac{\tan 3 A + \tan 3 B}{1 - \tan 3 Atan 3 B} = - \tan 3 C \\ \Rightarrow \tan 3 A + \tan 3 B + \tan 3 C = \tan 3 Atan 3 Btan 3 C \\ \Rightarrow \sum (\frac{3 \tan A - \tan^{3} A}{1 - 3 \tan^{2} A}) = \prod (\frac{3 \tan A - \tan^{3} A}{1 - 3 \tan^{2} A}) \\ \Rightarrow \sum (\frac{3 x - x^{3}}{1 - 3 x^{2}}) = \prod (\frac{3 x - x^{3}}{1 - 3 x^{2}}) \end{array}$