Factorise a3+8b3+27c2-18abc
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Given a3+27b3+8c3-18abc
= (a3+(3b)3+(2c)3-3*a*3b*2c) // use of the identity (a3+b3+c3-3abc) = (a+b+c) (a2+b2+c2-ab-bc-ca) //
= (a+3b+2c) (a2+(3b)2+(2c)2-a*3b-3b*2c-2c*a) // a=a,b= 3b, c= 2c //
= (a+3b+2c) (a2+9b2+4c2-3ab-6bc-2ca)
= (a3+(3b)3+(2c)3-3*a*3b*2c) // use of the identity (a3+b3+c3-3abc) = (a+b+c) (a2+b2+c2-ab-bc-ca) //
= (a+3b+2c) (a2+(3b)2+(2c)2-a*3b-3b*2c-2c*a) // a=a,b= 3b, c= 2c //
= (a+3b+2c) (a2+9b2+4c2-3ab-6bc-2ca)
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