Question

phan tich đa thức thành nhân tử(x^2-y^2) ^3+(y^2-z^2)^3+(z^2-x^2)^3 Giúp mk vơiz

Answer 1

Ta có (x^2 + y^2 )^3 + (z^2 – x^2 )^3 – (y^2 + z^2 )^3

= (x^2 + y^2 )^3 + (z^2 – x^2 )^3 + (-y^2 - z^2 )^3

Ta thấy x^2 + y^2 + z^2 – x^2 – y^2 – z^2 = 0

=> áp dụng nhận xét ta có: (x^2+y^2 )^3+ (z^2 -x^2 )^3 -y^2 -z^2 )^3

=3(x^2 + y^2 ) (z^2 –x^2 ) (-y^2 – z^2 )

= 3(x^2+y^2 ) (x+z)(x-z)(y^2+z^2 )

Answer 2

\((x^2+y^2)^3+(z^2-x^2)^3-(y^2+z^2)^3\)

\(=-3[x^4y^2-x^4z^2-x^2y^2z^2+x^2z^4-x^2y^4+x^2y^2z^2+y^4z^2-y^2z^4\)

\(=-3[x^2(x^2y^2-x^2z^2-z^2y^2+z^4)-y^2(x^2y^2-x^2z^2-z^2y^2+z^4)\)

\(=-3(x^2-y^2)(x^2y^2-x^2z^2-z^2y^2+z^4)\)

\(=-3(x^2-y^2[x^2(y^2-z^2)-z^2(y^2-z^2)]\)

\(=-3(x^2-y^2)(x^2-z^2)(y^2-z^2)\)

\(=-3(x-y)(x+y)(x-z)(x+z)(y+z)(y-z)\)