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Question
Factorise: 27x3+y3+z3−9xyz
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Solution
27x3+y3+z3−9xyz=(3x)3+(y)3+z3−3(3x)(y)(z)=(3x+y+z){(3x)2+(y)2+(z)2−(3x)(y)−(y)(z)−(z)(3x)}=(3x+y+z)(9x2+y2+z2−3xy−yz−3xz)
27x3+y3+z3−9xyz
=(3x)3+(y)3+z3−3(3x)(y)(z)
=(3x+y+z){(3x)2+(y)2+(z)2−(3x)(y)−(y)(z)−(z)(3x)}
=(3x+y+z)(9x2+y2+z2−3xy−yz−3xz)
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