How to factorise m³×n²-3m²×n³
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Answer:
(m^3-n^3) - m(m^2-n^2) + n(m-n)^3
= (m-n)(m^2+mn+n^2) - m(m+n)(m-n) + n(m-n)(m-n)(m-n)
= (m-n)[(m^2+mn+n^2) - m(m+n) + n(m-n)(m-n)]
= (m-n)[m^2+mn+n^2 - m^2-mn + n(m^2–2mn+n^2)]
= (m-n)[m^2+mn+n^2 - m^2-mn + m^2 n–2mn^2+n^3]
= (m-n)[n^2+m^2 n-2mn^2+n^3]
= n(m-n)[n + m^2 - 2mn + n^2]
= n(m-n)[n + (m-n)^2].
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