First slide

Theory of expressions

Question

If $α$ $β$ and $γ$ are the roots of the equation $x^{3} - 7 x + 7 = 0, then \frac{1}{α^{4}} + \frac{1}{β^{4}} + \frac{1}{γ^{4}}$ is

Moderate

A

7/3
B

3/7
C

4/7
D

7/4

Solution

Here, $Σα = 0, Σαβ = - 7, αβγ = - 7$
$∴ \frac{1}{α^{4}} + \frac{1}{β^{4}} + \frac{1}{γ^{4}} = \frac{α^{4} β^{4} + β^{4} γ^{4} + γ^{4} α^{4}}{α^{4} β^{4} γ^{4}}$
$= \frac{{Σα}^{4} β^{4}}{α^{4} β^{4} γ^{4}}$ …………………………..(i)
Now , $ΣαβΣαβΣαβΣαβ = (Σαβ)^{2} (Σαβ)^{2}$
$\begin{array}{r} \Rightarrow (- 7)^{4} = [α^{2} β^{2} + β^{2} γ^{2} + γ^{2} α^{2} + 2 αβγ (α + β + γ)] \\ [α^{2} β^{2} + β^{2} γ^{2} + γ^{2} α^{2} + 2 αβγ (α + β + γ)] \\ = (α^{2} β^{2} + β^{2} γ^{2} + γ^{2} α^{2}) (α^{2} β^{2} + β^{2} γ^{2} + γ^{2} α^{2}) \\ [∵ Σα = α + β + γ = 0] \end{array}$
$\begin{array}{l} = α^{4} β^{4} + β^{4} γ^{4} + γ^{4} α^{4} + 2 α^{4} β^{2} γ^{2} + 2 α^{2} β^{4} γ^{2} + 2 α^{2} β^{2} γ^{4} \\ = Σ α^{4} β^{4} + 2 α^{2} β^{2} γ^{2} (α^{2} + β^{2} + γ^{2}) \\ = Σ α^{4} β^{4} + 2 α^{2} β^{2} γ^{2} [(Σ α)^{2} - 2 Σ α β] \\ = Σ α^{4} β^{4} + 2 α^{2} β^{2} γ^{2} [0 - 2 \times (- 7)] \\ = Σ α^{4} β^{4} + 2 (- 7)^{2} (2 \times 7) \\ \Rightarrow Σ α^{4} β^{4} = (- 7)^{4} + 4 (- 7)^{3} \\ \Rightarrow Σ α^{4} β^{4} = (- 7)^{3} (- 7 + 4) = - 3 (- 7)^{3} \end{array}$
On putting this value in Eq. (i), we get
$\frac{1}{α^{4}} + \frac{1}{β^{4}} + \frac{1}{γ^{4}} = \frac{- 3 (- 7)^{3}}{(- 7)^{4}} = \frac{- 3}{- 7} = \frac{3}{7}$

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