a2+b2=69 ,a4+b4=3409 then a3+b3=
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Answer:
Step-by-step explanation:
given
a^2+b^2=69
(a+b)^2-2ab=69----(1)
a^4+b^4=3409
(a^2+b^2)^2-2a^2b^2=3409
69)^2-2a^2b^2=3409(substitute the value of a^2+b^2=69 )
4761-2a^2b^2=3409
2a^2b^2=4761-3409
2a^2b^2=1352
a^2b^2=1352/2
(ab)^2=676
ab=26----(2)
substitute (2) in (1)
(a+b)^2-2*26=69
(a+b)^2=69+52
a+b^2=121
a+b=11
a^3+b^3=(a+b)(a^2+b^2-ab)
=(11)(69-26)
=11*43
=473
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Step-by-step explanation:
anyhotty girl uf u bored so zoooomsid 560 571 3546 passwoordd == 123456
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