if a+b =s and a^3 +b^3 =t find a^2 +b^2 and a^4 +b^4 interms of s and t .


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Question Number 49642 by maxmathsup by imad last updated on 08/Dec/18

$i f a + b = s a n d a^{3} + b^{3} = t f i n d a^{2} + b^{2} a n d a^{4} + b^{4} i n t e r m s o f s a n d t .$
	Answered by mr W last updated on 08/Dec/18

	$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2}) = (a + b) [{(a + b)}^{2} - 3 a b]$ $t = s (s^{2} - 3 a b)$ $\Rightarrow a b = \frac{1}{3} (s^{2} - \frac{t}{s})$ $a^{2} + b^{2} = a^{2} + 2 a b + b^{2} - 2 a b = {(a + b)}^{2} - 2 a b$ $= s^{2} - \frac{2}{3} (s^{2} - \frac{t}{s}) = \frac{1}{3} (s^{2} + \frac{2 t}{s})$ $a^{4} + b^{4} = {(a^{2} + b^{2})}^{2} - 2 {(a b)}^{2} = \frac{1}{9} {(s^{2} + \frac{2 t}{s})}^{2} - \frac{2}{9} {(s^{2} - \frac{t}{s})}^{2}$ $= \frac{1}{9} (s^{4} + \frac{4 t^{2}}{s^{2}} + 2 s t) - \frac{2}{9} (s^{4} + \frac{t^{2}}{s^{2}} - 2 s t)$ $= \frac{1}{9} (\frac{2 t^{2}}{s^{2}} + 6 s t - s^{4})$
	Commented by maxmathsup by imad last updated on 08/Dec/18

	$t h a n k y o u s i r .$
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