factorise 54m^3 + 128n^3
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Step-by-step explanation:
54m^3 + 128n^3=2(27m^3 + 64n^3)
=2((3m)^3+(4n)^3)
=2((3m+4n)((3m)^2+(4n)^2-(3m)(4n))
=2(3m+4n)(9m^2+16n^2-12mn)
=2(3m+4n)(9m^2+16n^2+2(3m)(4n)-2(3m)(4n)-12mn)
=2(3m+4n)(9m^2+16n^2+24mn-36mn)
=2(3m+4n)((3m+4n)^2-(6rootmn)^2)
=2(3m+4n)(3m+4n+6rootmn)(3m+4n-6rootmn)
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Answer:
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