Let $ a, b $ and $ c $ be the three distinct non-zero real numbers satisfying the system of equations $ \frac{1}{a}+\frac{1}{a-1}+\frac{1}{a-2}=1 $, $ \frac{1}{b}+\frac{1}{b-1}+\frac{1}{b-2}=1 $ and $ \frac{1}{c}+\frac{1}{c-1}+\frac{1}{c-2}=1 $. Then, $ a b c $ is equal to

Question

Let $ a, b $ and $ c $ be the three distinct non-zero real numbers satisfying the system of equations $ \frac{1}{a}+\frac{1}{a-1}+\frac{1}{a-2}=1 $, $ \frac{1}{b}+\frac{1}{b-1}+\frac{1}{b-2}=1 $ and $ \frac{1}{c}+\frac{1}{c-1}+\frac{1}{c-2}=1 $. Then, $ a b c $ is equal to, $ 1 $, $ 2 $, $ 3 $, $ 4 $

MARKS App · Accepted Answer

Given, 1 a + 1 a - 1 + 1 a - 2 = 1 1 b + 1 b - 1 + 1 b - 2 = 1 1 c + 1 c - 1 + 1 c - 2 = 1 We can see that a , b and c satisfy the equation of the form given as, 1 x + 1 ( x - 1 ) + 1 x - 2 = 1 Thus, the roots of the above equation are a , b and c . ⇒ x ( x - 1 ) ( x - 2 ) = ( x - 1 ) ( x - 2 ) + x ( x - 1 ) + x ( x - 2 ) x 3 - 3 x 2 + 2 x = 3 x 2 - 6 x + 2 x 3 - 6 x 2 + 8 x - 2 = 0 The product of the roots of a b c is 2 .

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