Taking x=a,y=2b,z=−3c, this is in the form (x+y+z)3−x3−y3−z3 We use factorisation (x+y+z)3−x3−y3−z3=3(x+y)(y+z)(z+x) Putting back the values of x,y,z, we obtain (a+2b−3c)3−a3−8b3+27c3=3(a+2b)(2b−3c)(a−3c)
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Q1
Factorise: (a+2b−3c)3−a3−8b3+27c3.
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Q2
Factorise the following: (a−12b+13c)3−a3+18b3−127c3