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Question Number 191798 by Abdullahrussell last updated on 30/Apr/23

Commented by Frix last updated on 30/Apr/23

$Qu 191753; just add - 3 + 24 = 21$
Commented by BaliramKumar last updated on 01/May/23

$- 3 \times 24 = - 72$ 👇
	Answered by mr W last updated on 30/Apr/23

	$x^{2} y + y^{2} z + z^{2} x + {x y}^{2} + {y z}^{2} + {z x}^{2}$ $= x y (x + y) + y z (y + z) + z x (z + x)$ $= x y (x + y + z) + y z (y + z + x) + z x (z + x + y) - 3 x y z$ $= (x + y + z) (x y + y z + z x) - 3 x y z$ $= 6 \times 3 - 3 \times (- 1)$ $= 21 ✓$
	Answered by BaliramKumar last updated on 01/May/23
	$(x^2 y+y^2 z+z^2 x)(xy^2 +yz^2 +zx^2 ) = {x^4 yz+xy^4 z+xyz^4 }+{(xy)^3 +(yz)^3 +(zx)^3 }+{3(xyz)^2 } = {−x^3 −y^3 −z^3 }+{(xy+yz+zx)((xy+yz+zx)^2 −3(xy^2 z+xyz^2 +x^2 yz))+3x^2 y^2 z^2 }+{3} = −{(x+y+z)((x+y+z)^2 −3(xy+yz+zx))+3xyz}+{3(3^2 −3(−x−y−z))+3(−1)^2 } +{ 3} = −{6(6^2 −3×3)+3(−1)}+{3(9−3(−6))+3}+{3} = −{159}+{84}+{3} = determinant (((−72))) p+q = 21 pq = −72 s^2 −(p+q)s+(pq) = 0 s^2 −21s−72=0 s = −3 or 24$
	$(x^{2} y + y^{2} z + z^{2} x) ({x y}^{2} + {y z}^{2} + {z x}^{2})$ $= {x^{4} y z + {x y}^{4} z + {x y z}^{4}} + {{(x y)}^{3} + {(y z)}^{3} + {(z x)}^{3}} + {3 {(x y z)}^{2}}$ $= {- x^{3} - y^{3} - z^{3}} + {(x y + y z + z x) ({(x y + y z + z x)}^{2} - 3 ({x y}^{2} z + {x y z}^{2} + x^{2} y z)) + 3 x^{2} y^{2} z^{2}} + {3}$ $= - {(x + y + z) ({(x + y + z)}^{2} - 3 (x y + y z + z x)) + 3 x y z} + {3 (3^{2} - 3 (- x - y - z)) + 3 {(- 1)}^{2}} + {3}$ $= - {6 (6^{2} - 3 \times 3) + 3 (- 1)} + {3 (9 - 3 (- 6)) + 3} + {3}$ $= - {159} + {84} + {3}$ $= \begin{matrix} - 72 \end{matrix}$ $p + q = 21 p q = - 72$ $s^{2} - (p + q) s + (p q) = 0$ $s^{2} - 21 s - 72 = 0$ $s = - 3 o r 24$
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