Question

Представьте в виде произведения: 1) m4+mn3-m3n-n4; 2) 8а3+6а2+3а+1

Answer 1

$1) m^4+mn^3-m^3n-n^4=(m^4-n^4)+(mn^3-m^3n)= =(m^2-n^2)(m^2+n^2)+mn(m^2-n^2)=(m^2-n^2)(m^2+mn+n^2) 2)8a^3+6a^2+3a+1=(8a^3+1)+(6a^2+3a)= =((2a)^3+1^3)+3a(2a+1)=(2a+1)(a^2-ab+b^2)+3a(2a+1)= =(2a+1)(a^2+3a-ab+b^2)$

Answer 2

1. m4 + mn3 - m3n - n4 can be factored as follows:

m4 + mn3 - m3n - n4 = m(m3 - n3) - n(m3 - n3) = (m - n)(m3 - n3)

Therefore, the expression can be written as (m - n)(m3 - n3).

2. 8a3 + 6a2 + 3a + 1 cannot be factored further using simple algebraic techniques.