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Question Number 192173 by universe last updated on 10/May/23
prove that  (x^2 +a^2 )^4  = (x^4 −6x^2 a^2 +a^4 )^2 +(4x^3 a−4xa^3 )^2
$${prove}\:{that} \\ $$$$\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{4}} \:=\:\left({x}^{\mathrm{4}} −\mathrm{6}{x}^{\mathrm{2}} {a}^{\mathrm{2}} +{a}^{\mathrm{4}} \right)^{\mathrm{2}} +\left(\mathrm{4}{x}^{\mathrm{3}} {a}−\mathrm{4}{xa}^{\mathrm{3}} \right)^{\mathrm{2}} \\ $$
Answered by mehdee42 last updated on 10/May/23
right side=((x^2 −a^2 )^2 −4a^2 x^2 )^2 +16a^2 x^2 (x^2 −a^2 )^2   (x^2 −a^2 )^4 −8a^2 x^2 (x^2 −a^2 )^2 +16a^2 x^2 +16a^2 x^2 (x^2 −a^2 )^2   (x^2 −a^2 )^4 +8a^2 x^2 (x^2 −a^2 )^2 +16a^2 x^2   ((x^2 −a^2 )^2  +4a^2 x^2 )^2 =(x^2 +a^2 )^2  ✓
$${right}\:{side}=\left(\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{16}{a}^{\mathrm{2}} {x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{4}} −\mathrm{8}{a}^{\mathrm{2}} {x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{16}{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{16}{a}^{\mathrm{2}} {x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{4}} +\mathrm{8}{a}^{\mathrm{2}} {x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{16}{a}^{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$$\left(\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\mathrm{4}{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)^{\mathrm{2}} =\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} \:\checkmark \\ $$$$ \\ $$

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