Question

Let \[\alpha \] , \[\beta \] , \[\gamma \] be three numbers such that \[2(\alpha \beta + \beta \gamma + \alpha \gamma ) = \alpha \beta \gamma \] , \[\dfrac{1}{{{\alpha ^2}}} + \dfrac{1}{{{\beta ^2}}} + \dfrac{1}{{{\gamma ^2}}} = \dfrac{9}{4}\] and \[\alpha + \beta + \gamma = 2\] then the value of \[\alpha \beta \gamma \] is
a) -4
b) -2
c) 2
d) 4

Answer 1

Hint: In the given question we are asked to find the value of \[\alpha \beta \gamma \] . We are also given a few results in the question that will help to find the value of \[\alpha \beta \gamma \] .In order to solve the question we will use the algebraic identity of the sum of three variables whole squared.

Complete step by step solution:
Given that,
 \[2(\alpha \beta + \beta \gamma + \alpha \gamma ) = \alpha \beta \gamma - - - - - (1)\]
 \[\dfrac{1}{{{\alpha ^2}}} + \dfrac{1}{{{\beta ^2}}} + \dfrac{1}{{{\gamma ^2}}} = \dfrac{9}{4} - - - - - (2)\]
And \[\alpha + \beta + \gamma = 2 - - - - - (3)\]
According to the algebraic identities we know that,
 \[[(a + b + c)^2] = {a^2} + {b^2} + {c^2} + 2(ab + bc + ca) - - - - - (4)\]
Let \[a = \dfrac{1}{\alpha }\] , \[b = \dfrac{1}{\beta }\] and \[c = \dfrac{1}{\gamma }\]
Now, putting the above values in (4)
We get,
 \[{\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta } + \dfrac{1}{\gamma }} \right)^2} = \dfrac{1}{{{\alpha ^2}}} + \dfrac{1}{{{\beta ^2}}} + \dfrac{1}{{{\gamma ^2}}} + 2\left( {\dfrac{1}{{\alpha \beta }} + \dfrac{1}{{\beta \gamma }} + \dfrac{1}{{\alpha \gamma }}} \right) - - - - - (5)\]
Taking LCM, we get,
 \[{\left( {\dfrac{{\beta \gamma + \alpha \gamma + \alpha \beta }}{{\alpha \beta \gamma }}} \right)^2} = \dfrac{1}{{{\alpha ^2}}} + \dfrac{1}{{{\beta ^2}}} + \dfrac{1}{{{\gamma ^2}}} + 2\left( {\dfrac{{\alpha + \beta + \gamma }}{{\alpha \beta \gamma }}} \right)\]
Now from (1) we can say that, \[\alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{{\alpha \beta \gamma }}{2}\]
And from (3) we can say that, \[\alpha + \beta + \gamma = 2\]
And that \[\dfrac{1}{{{\alpha ^2}}} + \dfrac{1}{{{\beta ^2}}} + \dfrac{1}{{{\gamma ^2}}} = \dfrac{9}{4}\] from (2)
Therefore, putting all of the above values in equation (5)
We get,
 \[{\left( {\dfrac{{\alpha \beta \gamma }}{2}} \right)^2} \times \dfrac{1}{{{{\left( {\alpha \beta \gamma } \right)}^2}}} = \dfrac{9}{4} + 2\left( {\dfrac{2}{{\alpha \beta \gamma }}} \right)\]
Eliminating common factors, we get,
 \[\dfrac{1}{4} = \dfrac{9}{4} + \dfrac{4}{{\alpha \beta \gamma }}\]
 \[ \Rightarrow \dfrac{1}{4} - \dfrac{9}{4} = \dfrac{4}{{\alpha \beta \gamma }}\]
Therefore,
 \[ \Rightarrow - \dfrac{8}{4} = \dfrac{4}{{\alpha \beta \gamma }}\]
 \[ \Rightarrow - 2 = \dfrac{4}{{\alpha \beta \gamma }}\]
Thus,
 \[ \Rightarrow \alpha \beta \gamma = - \dfrac{4}{2}\]
Hence,
 \[ \Rightarrow \alpha \beta \gamma = - 2\] which is our required solution.
So, the correct answer is “Option B”.

Note: Algebraic identities are always helpful in solving questions related to polynomials like we have used. In the given question. One of the best ways to solve a question related to polynomials and identities is through substitution. Substitution means the substitution of any variable value on both sides of the algebraic identity. If you have correctly expanded or fixed an example with help of algebraic identities, then any variable value will hold true for both the left and the right sides of the equation.