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The Wonders of Arithmetic from Pierre Simon de Fermat
Год написания книги
2021
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=x
zz
+y(z
+xz
)+…
(x±y)
=z
=x
±y(x
+x
z+x
z
+…+xz
+z
) (7)
We will name the expression in brackets consisting of n summand a symmetric polynomial and we will present it in the form (x ++ z)
as an abridged spelling. Now using formula (7), we will exponentiating n both sides of formula (2) as follows.
[a?(c?b)]
=a
+{b
?c
+(c
?b
)}?(c?b)[a
+a
2m+…
+ a(2m)
+(2m)
]=(2m)
Now through identity
(c
?b
)=(c?b)(c
+c
b+…+cb
+b
) we obtain:
{a
+b
?c
}+(c?b)[(c++b)
?(a++2m)
]=(2m)
(8)
Equation (8) is a formula (2) raised to the power n what can be seen after substituting c?b=a?2m in (8) and obtaining the identity[59 - In this case, identity (9) indicates that the same key formula is substituted into the transformed key formula (2) or that the equation (8) we obtained, is a key formula (2) in power n. But you can go the reverse way just give the identity (9) and then divide into factor the differences of powers and such a way you can obtain (8) without using the Fermat Binominal (7). But this way can be a trick to hide the understanding of the essence because when some identity falls from the sky, it may seem that there is nothing to object. However, if you memorize only this path, there is a risk of exposure in a misunderstanding of the essence because the question how to obtain this identity, may go unanswered.]:
{a
+b
?c
}+(c
?b
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