if x+y+z=1 x^2 +y^2 +z^2 =2 x^3 +y^3 +z^3 =3 find x^4 +y^4 +z^4


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Question Number 52592 by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

$i f x + y + z = 1$ $x^{2} + y^{2} + z^{2} = 2$ $x^{3} + y^{3} + z^{3} = 3$ $f i n d x^{4} + y^{4} + z^{4}$
	Answered by math1967 last updated on 10/Jan/19

	$x + y + z = 1$ $\Rightarrow 2 (x y + y z + z x) = 1 - 2 = - 1$ $∴ (x y + y z + z x) = - \frac{1}{2}$ $x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} + 2 x y z (x + y + z) = \frac{1}{4}$ $x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} = \frac{1}{4} - \frac{1}{3} = - \frac{1}{12} ★$ $x^{3} + y^{3} + z^{3} - 3 x y z + 3 x y z = 3 ★$ $(x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) + 3 x y z = 3$ $3 x y z = 3 - \frac{5}{2} = \frac{1}{2} \Rightarrow x y z = \frac{1}{6}$ ${(x^{2} + y^{2} + z^{2})}^{2} = 4$ $x^{4} + y^{4} + z^{4} + 2 (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2}) = 4$ $x^{4} + y^{4} + z^{4} = 4 + \frac{1}{6} = 4 \frac{1}{6} a n s$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

	$b a h d a r u n e x c e l l e n t . . .$
	Commented by math1967 last updated on 10/Jan/19

	$d h a n y a b a d$
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