Question

Factorise (27x)³+y³+z³-9xyz

Answer 1

(27x)³+y³+z³-9xyz

=(3x)³+y³+z³-3(3x)(y) (z)

=(3x+y+z) [(3x)²+y²+z²-(3x)(y) -(y)(z) -(z)(3x) ]

=(3x+y+z) (9x²+y²+z²-3xy-yz-3zx)

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