Find the sum of the fourth powers of the roots of equation: 7x^3 −21x^2 +9x+2=0


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Question Number 198919 by necx122 last updated on 25/Oct/23

$F i n d t h e s u m o f t h e f o u r t h p o w e r s o f$ $t h e r o o t s o f e q u a t i o n :$ $7 x^{3} - 21 x^{2} + 9 x + 2 = 0$
Commented by necx122 last updated on 25/Oct/23
wow! Really learning new things. Thank you sir.
	Answered by deleteduser1 last updated on 25/Oct/23

	$L e t r o o t s b e a, b, c, . T h e n a + b + c = \frac{- (- 21)}{7} = 3$ $a b + b c + c a = \frac{9}{7}; a b c = \frac{- 2}{7}$ $a^{4} + b^{4} + c^{4} = {(a^{2} + b^{2} + c^{2})}^{2} - 2 [{(a b)}^{2} + {(b c)}^{2} + {(c a)}^{2}]$ $(a^{2} + b^{2} + c^{2}) = {(a + b + c)}^{2} - 2 (a b + b c + c a) = 9 - \frac{18}{7}$ $= \frac{63 - 18}{7} = \frac{45}{7}$ ${(a b)}^{2} + {(b c)}^{2} + {(c a)}^{2} = {[a b + b c + c a]}^{2} - 2 a b c (a + b + c)$ $= \frac{81}{49} + \frac{84}{49} = \frac{165}{49}$ $\Rightarrow a^{4} + b^{4} + c^{4} = {(\frac{45}{7})}^{2} - \frac{330}{49} = \frac{1695}{49}$
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