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Factorize-2x-2-y-2-2y-2-z-2-2z-2-x-2-x-4-y-4-z-4-

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Question Number 174324 by BaliramKumar last updated on 29/Jul/22
Factorize:− 2x^2 y^2 + 2y^2 z^2 + 2z^2 x^2 − x^4 − y^4 − z^4
$$\mathrm{F}{actorize}:− \\ $$$$\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:+\:\mathrm{2}{y}^{\mathrm{2}} {z}^{\mathrm{2}} \:+\:\mathrm{2}{z}^{\mathrm{2}} {x}^{\mathrm{2}} \:−\:{x}^{\mathrm{4}} \:−\:{y}^{\mathrm{4}} \:−\:{z}^{\mathrm{4}} \\ $$
Commented by kaivan.ahmadi last updated on 30/Jul/22
−(x^2 −y^2 −z^2 )^2 +4y^2 z^2 = (2yz−x^2 +y^2 +z^2 )(2yz+x^2 −y^2 −z^2 )= ((y+z)^2 −x^2 )(x^2 −(y+z)^2 )= (y+z−x)(y+z+x)(x+y+z)(x−y−z)= −(x+y+z)^2 (x−y−z)^2
$$−\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} −{z}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} {z}^{\mathrm{2}} = \\ $$$$\left(\mathrm{2}{yz}−{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left(\mathrm{2}{yz}+{x}^{\mathrm{2}} −{y}^{\mathrm{2}} −{z}^{\mathrm{2}} \right)= \\ $$$$\left(\left({y}+{z}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −\left({y}+{z}\right)^{\mathrm{2}} \right)= \\ $$$$\left({y}+{z}−{x}\right)\left({y}+{z}+{x}\right)\left({x}+{y}+{z}\right)\left({x}−{y}−{z}\right)= \\ $$$$−\left({x}+{y}+{z}\right)^{\mathrm{2}} \left({x}−{y}−{z}\right)^{\mathrm{2}} \\ $$
Commented by BaliramKumar last updated on 30/Jul/22
recheck answer
$${recheck}\:{answer}\: \\ $$
Commented by dragan91 last updated on 30/Jul/22
−(x^2 −y^2 −z^2 )^2 +4y^2 z^2 = (2yz−x^2 +y^2 +z^2 )(2yz+x^2 −y^2 −z^2 )= ((y+z)^2 −x^2 )(x^2 −(y−z)^2 )= (y+z−x)(y+z+x)(x+y−z)(x−y+z)= −(x+y+z)^2 (x−y−z)^2
$$−\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} −{z}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} {z}^{\mathrm{2}} = \\ $$$$\left(\mathrm{2}{yz}−{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left(\mathrm{2}{yz}+{x}^{\mathrm{2}} −{y}^{\mathrm{2}} −{z}^{\mathrm{2}} \right)= \\ $$$$\left(\left({y}+{z}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −\left({y}−{z}\right)^{\mathrm{2}} \right)= \\ $$$$\left({y}+{z}−{x}\right)\left({y}+{z}+{x}\right)\left({x}+{y}−{z}\right)\left({x}−{y}+{z}\right)= \\ $$$$\cancel{−\left({x}+{y}+{z}\right)^{\mathrm{2}} \left({x}−{y}−{z}\right)^{\mathrm{2}} } \\ $$
Commented by Frix last updated on 30/Jul/22
I know this from Heron′s Triangle Formula the area of a triangle with sides a, b, c is A=((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/4)= =((√(2(a^2 b^2 +b^2 c^2 +c^2 a^2 )−(a^4 +b^4 +c^4 )))/4)
$$\mathrm{I}\:\mathrm{know}\:\mathrm{this}\:\mathrm{from}\:\mathrm{Heron}'\mathrm{s}\:\mathrm{Triangle}\:\mathrm{Formula} \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{sides}\:{a},\:{b},\:{c}\:\mathrm{is} \\ $$$${A}=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}}= \\ $$$$\:\:\:\:\:=\frac{\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)−\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)}}{\mathrm{4}} \\ $$
Commented by BaliramKumar last updated on 30/Jul/22
Yes sir
$${Yes}\:{sir} \\ $$
Answered by dragan91 last updated on 30/Jul/22
easy
$$\mathrm{easy} \\ $$
Answered by dragan91 last updated on 30/Jul/22
−((x^2 −y^2 −z^2 )^2 −4y^2 z^2 )= −(x^2 −y^2 −2yz−z^2 )(x^2 −y^2 +2yz−z^2 )= −(x^2 −(y+z)^2 )(x^2 −(y−z)^2 )= −(x−y−z)(x+y+z)(x−y+z)(x+y−z)= (−x+y+z)(x+y+z)(x−y+z)(x+y−z)
$$−\left(\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} −\mathrm{z}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4y}^{\mathrm{2}} \mathrm{z}^{\mathrm{2}} \right)= \\ $$$$−\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} −\mathrm{2yz}−\mathrm{z}^{\mathrm{2}} \right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} +\mathrm{2yz}−\mathrm{z}^{\mathrm{2}} \right)= \\ $$$$−\left(\mathrm{x}^{\mathrm{2}} −\left(\mathrm{y}+\mathrm{z}\right)^{\mathrm{2}} \right)\left(\mathrm{x}^{\mathrm{2}} −\left(\mathrm{y}−\mathrm{z}\right)^{\mathrm{2}} \right)= \\ $$$$−\left(\mathrm{x}−\mathrm{y}−\mathrm{z}\right)\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left(\mathrm{x}−\mathrm{y}+\mathrm{z}\right)\left(\mathrm{x}+\mathrm{y}−\mathrm{z}\right)= \\ $$$$\left(−\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left(\mathrm{x}−\mathrm{y}+\mathrm{z}\right)\left(\mathrm{x}+\mathrm{y}−\mathrm{z}\right) \\ $$$$ \\ $$$$ \\ $$

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