Question

If $\alpha ,\beta ,\gamma $ are such that $\alpha + \beta + \gamma = 2$ , ${\alpha ^2} + {\beta ^2} + {\gamma ^2} = 6$, ${\alpha ^3} + {\beta ^3} + {\gamma ^3} = 8$ , then ${\alpha ^4} + {\beta ^4} + {\gamma ^4}$ is equal to
(A) $18$
(B) $10$
(C) $15$
(D) $36$

Answer 1

Hint:
Start with using the identity ${\left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2}} \right)^2} = {\alpha ^4} + {\beta ^4} + {\gamma ^4} + 2{\alpha ^2}{\beta ^2} + 2{\beta ^2}{\gamma ^2} + 2{\gamma ^2}{\alpha ^2}$ and substitute the known value in it. Now use identity ${\left( {\alpha + \beta + \gamma } \right)^2} = {\alpha ^2} + {\beta ^2} + {\gamma ^2} + 2\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right)$ to find the value of $\alpha \beta + \beta \gamma + \gamma \alpha $ . Then use \[{\alpha ^3} + {\beta ^3} + {\gamma ^3} - 3\alpha \beta \gamma = \left( {\alpha + \beta + \gamma } \right)\left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2} - \alpha \beta - \beta \gamma - \gamma \alpha } \right)\] to find the value of \[\alpha \beta \gamma \] . Now by using \[{\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right)^2} = {\alpha ^2}{\beta ^2} + {\beta ^2}{\gamma ^2} + {\gamma ^2}{\alpha ^2} + 2\alpha \beta \gamma \left( {\alpha + \beta + \gamma } \right)\] we can find the value of \[{\alpha ^2}{\beta ^2} + {\beta ^2}{\gamma ^2} + {\gamma ^2}{\alpha ^2}\]. Now substitute this value in the initial relation to get the answer.

Complete step by step solution:
Here we are given with three equations $\alpha + \beta + \gamma = 2$ , ${\alpha ^2} + {\beta ^2} + {\gamma ^2} = 6$ and ${\alpha ^3} + {\beta ^3} + {\gamma ^3} = 8$ in term of $\alpha ,\beta ,\gamma $ . With these given expressions, we need to find the value of ${\alpha ^4} + {\beta ^4} + {\gamma ^4}$
As we know the following identity: ${\left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2}} \right)^2} = {\alpha ^4} + {\beta ^4} + {\gamma ^4} + 2{\alpha ^2}{\beta ^2} + 2{\beta ^2}{\gamma ^2} + 2{\gamma ^2}{\alpha ^2}$
But we already have the value ${\alpha ^2} + {\beta ^2} + {\gamma ^2} = 6$ , substituting this into the above equation, we get:
$ \Rightarrow {\left( 6 \right)^2} = {\alpha ^4} + {\beta ^4} + {\gamma ^4} + 2{\alpha ^2}{\beta ^2} + 2{\beta ^2}{\gamma ^2} + 2{\gamma ^2}{\alpha ^2}$
$ \Rightarrow {\alpha ^4} + {\beta ^4} + {\gamma ^4} + 2\left( {{\alpha ^2}{\beta ^2} + {\beta ^2}{\gamma ^2} + {\gamma ^2}{\alpha ^2}} \right) = 36$ ........(i)
So to find the required value for ${\alpha ^4} + {\beta ^4} + {\gamma ^4}$ , we need to solve the above equation and for that, we need to have the value for ${\alpha ^2}{\beta ^2} + {\beta ^2}{\gamma ^2} + {\gamma ^2}{\alpha ^2}$ .
We know the identity ${\left( {\alpha + \beta + \gamma } \right)^2} = {\alpha ^2} + {\beta ^2} + {\gamma ^2} + 2\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right)$
Substituting the known values of $\alpha + \beta + \gamma $ and ${\alpha ^2} + {\beta ^2} + {\gamma ^2}$
$ \Rightarrow {\left( {\alpha + \beta + \gamma } \right)^2} = {\alpha ^2} + {\beta ^2} + {\gamma ^2} + 2\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) \Rightarrow {\left( 2 \right)^2} = 6 + 2\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right)$
On further simplifying this, we get:
$ \Rightarrow 4 = 6 + 2\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) \Rightarrow \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{{ - 2}}{2} = - 1$ ........(ii)
Now using the identity:
\[{\alpha ^3} + {\beta ^3} + {\gamma ^3} - 3\alpha \beta \gamma = \left( {\alpha + \beta + \gamma } \right)\left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2} - \alpha \beta - \beta \gamma - \gamma \alpha } \right)\]
Substituting the known values from (ii) and given expressions, we get:
\[ \Rightarrow 8 - 3\alpha \beta \gamma = \left( 2 \right)\left( {6 - \left( { - 1} \right)} \right) = 2 \times 7 = 14\]
On further simplifying this, we can write it as:
\[ \Rightarrow 8 - 3\alpha \beta \gamma = 14 \Rightarrow - 3\alpha \beta \gamma = 6 \Rightarrow \alpha \beta \gamma = - 2\] .........(iii)
Now, using the identity: ${\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right)^2} = {\alpha ^2}{\beta ^2} + {\beta ^2}{\gamma ^2} + {\gamma ^2}{\alpha ^2} + 2\alpha {\beta ^2}\gamma + 2\alpha \beta {\gamma ^2} + 2{\alpha ^2}\beta \gamma $ and substituting the known values in the expression to get:
\[ \Rightarrow {\left( { - 1} \right)^2} = {\alpha ^2}{\beta ^2} + {\beta ^2}{\gamma ^2} + {\gamma ^2}{\alpha ^2} + 2\alpha \beta \gamma \left( {\alpha + \beta + \gamma } \right) = {\alpha ^2}{\beta ^2} + {\beta ^2}{\gamma ^2} + {\gamma ^2}{\alpha ^2} + 2\left( { - 2} \right)\left( 2 \right)\]
 We can further simplify this equation as:
\[ \Rightarrow {\left( { - 1} \right)^2} = {\alpha ^2}{\beta ^2} + {\beta ^2}{\gamma ^2} + {\gamma ^2}{\alpha ^2} + 2\left( { - 2} \right)\left( 2 \right) \Rightarrow {\alpha ^2}{\beta ^2} + {\beta ^2}{\gamma ^2} + {\gamma ^2}{\alpha ^2} = 1 + 8 = 9\]
Therefore, we get: \[{\alpha ^2}{\beta ^2} + {\beta ^2}{\gamma ^2} + {\gamma ^2}{\alpha ^2} = 9\] ...........(iv)
Now let’s go back to our relation (i) and substitute the value from (iv) in it to get the required value:
 $ \Rightarrow {\alpha ^4} + {\beta ^4} + {\gamma ^4} + 2\left( {{\alpha ^2}{\beta ^2} + {\beta ^2}{\gamma ^2} + {\gamma ^2}{\alpha ^2}} \right) = 36 \Rightarrow {\alpha ^4} + {\beta ^4} + {\gamma ^4} + 2\left( 9 \right) = 36$
Now we can transpose the constants to the right side to get the required value of the expression:
\[ \Rightarrow {\alpha ^4} + {\beta ^4} + {\gamma ^4} + 18 = 36 \Rightarrow {\alpha ^4} + {\beta ^4} + {\gamma ^4} = 36 - 18 = 18\]
Therefore, we get the required value of ${\alpha ^4} + {\beta ^4} + {\gamma ^4}$ as $18$

Hence, the option (A) is the correct answer.

Note:
In questions like this, you should start with using the algebraic identities to form a relationship in the required expression. Here the use of correct identity can make the solution less complicated. An alternative approach to this problem can be to evaluate relation (ii), (iii) and (iv) before substituting it in the relation (i). Go step by step for making process less complicated.