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# Answer to Question #11212 in Algebra for n k siva prasad

Question #11212
factorize

m^8-11m^4n^4-80n^8
1
2012-06-22T08:55:40-0400
A = m^8-11m^4n^4-80n^8
= n^8 ( (m/n)^8 - 11 (m/n)^4 - 80 )

Replace: t=(m/n)^4.
Then

A = n^8 ( t^2 - 11t - 80
)

Let us solve
It is easy to verify that the equation:
t^2 - 11t
- 80 = 0
has the following two roots:
t1 = 16
t2 = -5.
Hence
t^2
- 11t - 80 = (t-16)(t+5).

Thus

A = n^8 (t-16)(t+5)
= n^8 *
( (m/n)^4 - 16 ) * ( (m/n)^4 + 5 )
= ( m^4 - 16 n^4 ) * ( m^4 + 5 n^4 ) =

= ( (m^2)^2 - (4n^2)^2 ) * ( m^4 + 5 n^4 ) =
= (m^2-4n^2) * (m^2 +
4n^2) * ( m^4 + 5 n^4 ) =
= ( m^2 - (2n)^2 ) * (m^2 + 4n^2) * ( m^4 + 5
n^4 ) =
= (m-2n) * (m+2n) * (m^2 + 4n^2) * ( m^4 + 5 n^4
).

Thus
m^8 - 11m^4n^4 - 80n^8 = (m-2n) * (m+2n) * (m^2 +
4n^2) * ( m^4 + 5 n^4 ).

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