If a-b-c-abc- show that a b 2 c 2 -1 bc-1 - b c 2 a 2 -1 ca-1 - c a 2 b 2 -1 ab-1 -2abc

L.H.S. = a(bc 1)(bc+1) bc+1 + b(ca 1)(ca+1) ca+1 + c(ab 1)(ab+1) ab+1 =a(bc 1)+b(ca 1)+c(ab 1) =abc a+abc b+abc c =3abc a b c =3ab ...

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Arithmetic Progression

If a+b+c=ab...

Question

If $a + b + c = a b c$ , show that
$\frac{a (b^{2} c^{2} - 1)}{b c + 1} + \frac{b (c^{2} a^{2} - 1)}{c a + 1} + \frac{c (a^{2} b^{2} - 1)}{a b + 1} = 2 a b c$

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Solution

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Similar questions

If a + b + c = 0, then prove the following

(a) (b + c) (b − c) + a(a + 2b) = 0

(b) a(a² − bc) + b(b² − ca) + c(c² − ab) = 0

(d)

\frac{a (b^{2} + c^{2}) + b (c^{2} + a^{2}) + c (a^{2} + b^{2})}{a b c}

a b + b c + c a = 0

\frac{1}{a^{2} - b c} + \frac{1}{b^{2} - c a} + \frac{1}{c^{2} - a b} = 0

a^{2} + b^{2} + c^{2} = 1

a b + b c + c a > - \frac{1}{2}

△ A B C, ∠ C = 90^{o}

p

C

A B

A B = c, B C = a, C A = b

\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}

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