expand α^4 +β^(β ) please


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Question Number 33515 by alekan251 last updated on 18/Apr/18

$e x p a n d α^{4} + β^{β} p l e a s e$
Commented by math khazana by abdo last updated on 18/Apr/18

$i f y o u m e a n α^{4} + β^{4} f r o m R$ $= {(α^{2})}^{2} + {(β^{2})}^{2} = {(α^{2} + β^{2})}^{2} - 2 α^{2} β^{2}$ $= (α^{2} + β^{2} - \sqrt{2} α β) (α^{2} + β^{2} + \sqrt{2} α β)$ $i f α a n d β f r o m R l e t f i x β a n d t h e r o o t s o f$ $α^{2} - \sqrt{2} α β + β^{2}$ $Δ = {(- \sqrt{2} β)}^{2} - 4 β^{2} = - 2 β^{2} = {(i \sqrt{2} β)}^{2}$ $α_{1} = \frac{\sqrt{2} β + i \sqrt{2} β}{2} = \frac{\sqrt{2} β}{2} (1 + i)$ $α_{2} = \frac{\sqrt{2} β - i \sqrt{2} β}{2} = \frac{\sqrt{2} β}{2} (1 - i) a l s o w e h a v e$ $α^{2} + \sqrt{2} α β + β^{2} = α^{2} - \sqrt{2} α (- β) + {(- β)}^{2} s o t h e$ $r o o t s a r e t_{1} = - \frac{β \sqrt{2}}{2} (1 + i) a n d t_{2} = - \frac{β \sqrt{2}}{2} (1 - i)$ $α^{4} + β^{4} = (α - α_{1}) (α - α_{2}) (α - t_{1}) (α - t_{2}) .$
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