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Miscellaneous

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Last updated at May 29, 2018 by Teachoo

Misc 19 If (π+ππ)(π+ππ)(π+ππ)(π+πβ)=π΄+ππ΅, then show that (π2 + π2) (π2 + π2) (π2 + π2) (π2 + β2) = π΄2 +π΅2. Introduction (π΄ + ππ΅) ( π΄ β ππ΅) Using ( a β b ) ( a + b ) = a2 β b2 = π΄2 β (ππ΅)2 = π΄2 β π2 π΅2 Putting i2 = β1 = π΄2 β ( β1) π΅2 = π΄2 +π΅2 Hence, (π΄ + ππ΅) (π΄ β ππ΅) = π΄2 +π΅2 Misc, 19 If (π+ππ)(π+ππ)(π+ππ)(π+πβ)=π΄+ππ΅, then show that (π2 + π2) (π2 + π2) (π2 + π2) (π2 + β2) = π΄2 +π΅2. Given ( π΄ + ππ΅ ) = (π + ππ ) ( π + ππ ) (π + ππ ) ( π + πβ ) To calculate ( π΄ β ππ΅ ) Replacing π by βπ in (1) (π΄ βππ΅ ) = ( π β ππ ) ( π β ππ ) ( π β ππ ) ( π β πβ ) Now, calculating (π΄ + ππ΅) ( π΄ β ππ΅) (π΄ + ππ΅) ( π΄ β ππ΅) = (π + ππ )( π + ππ )(π + ππ )( π + πβ )(π β ππ ) ( π β ππ ) (π β ππ ) ( π β πβ ) π΄2 + π΅^2= [( π+ ππ )(π β ππ )][(π+ ππ)(π β ππ )] [( π + ππ) ( π β ππ )] [( π + πβ ) ( π β πβ)] ππ πππ ( π₯ β π¦ ) ( π₯ + π¦ ) = π₯2+π¦2 = [(π)^2 β (ππ)2] [ π2 β ( ππ)^2] [π2β (ππ)^2 ] [π2 β (β πβ)]2 = [ π2 β π2 π2 ] [ π2 β π2 π2 ] [ π2 β π2 π2 ] [ π2 β π2 β2 ] Putting i2 = β1 = [ π2β (β1)π2] [ π2 β (β1) π ] [ π2 β (β1) π)] [π2 β (β1) β2 ] = [ π2 + π2 ] [ π2 + π2 ] [ π2 + π2 ] [π2 + β2 ] Hence, (π2 + π2) (π2 + π2) (π2 + π2) (π2 + β2) = π΄2 +π΅2. Hence proved