alpha and beata??? help please

lorr3

formula for alpha^3+beata^3 ... is it (a+b)^3 -2ab(a+b)??

likely_lass

-3ab(a+b)

If you want to learn it off, it's:

a^3 + b^3 = (a+b)^3 - 3ab(a+b)

Derivation:

(a+b)^3 = (a^2 + 2ab + b^2)(a+b)
= a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3
= a^3 + b^3 = (a+b)^3 - 2ab(a+b) -ab(a+b)
= a^3 + b^3 = (a+b)^3 - 3ab(a+b)

lorr3

thank you

papu

-JammyDodger- wrote: »

If you want to learn it off, it's:

a^3 + b^3 = (a+b)^3 - 3ab(a+b)

Derivation:

(a+b)^3 = (a^2 + 2ab + b^2)(a+b)
= a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3
= a^3 + b^3 = (a+b)^3 - 2ab(a+b) -ab(a+b)
= a^3 + b^3 = (a+b)^3 - 3ab(a+b)

bit different but the same , just incase it didnt make sense to anyone

zodac

Can (a^4 + b^4) be asked? It's in the notes in the past papers but not in the maths book itself. I'm not sure if it on the syllabus or not...

Zonda999

zodac wrote: »

Can (a^4 + b^4) be asked? It's in the notes in the past papers but not in the maths book itself. I'm not sure if it on the syllabus or not...

Couldnt see it there really. I imagine a lot of people wouldnt know a^3 +b^3 so it would be extremely difficult to work with a^4 + b^4

likely_lass

just work out (a+b)(a+b)(a+b)(a+b) thats (a+b)^4 then subtract whatever you need to get just (a^4 + b^4)

dont think its very likely