Question

If $α, β, γ$ are such that $α+β+γ=2$, $α²+β²+γ²=6$, $α³+β³+γ³=8$, then $α⁴+β⁴+γ⁴$ is

• A. 5
• B. 18
• C. 12
• D. 36

Hint:

If $α, β, γ$ are such that $α+β+γ=2$, $α+β+γ=6$, $α+β+γ=8$, then $α+β+γ$ is

Answer 1

The Correct Option is B

Step 1: Concept
Use algebraic identities for sums of powers.
Step 2: Analysis
$(\alpha+\beta+\gamma)^2 = \alpha^2+\beta^2+\gamma^2 + 2(\alpha\beta+\beta\gamma+\gamma\alpha)$. Substituting values: $4 = 6 + 2(\sum \alpha\beta) \Rightarrow \sum \alpha\beta = -1$.
Step 3: Evaluation
Using $\alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\sum \alpha)(\sum \alpha^2 - \sum \alpha\beta)$, we get $8 - 3\alpha\beta\gamma = 2(6 - (-1)) = 14 \Rightarrow \alpha\beta\gamma = -2$.
Step 4: Conclusion
$(\sum \alpha^2)^2 = \sum \alpha^4 + 2\sum \alpha^2\beta^2$. After further calculation involving $\sum \alpha\beta$ and $\alpha\beta\gamma$, $\alpha^4+\beta^4+\gamma^4$ is found to be 18.
Final Answer: (b)