Question

factorise: 27x^3+y^3+z^3-9xyz​

Answer 1

Answer:

27x3+y3+z3-9xyz= (3x)3+(y)3+(z)3-3(3x)(y)(z)

by identity = a3 + b3 +c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ac)

(3x+y+z)((3x)2+(y)2+(z)2-3xy-yz-3xz

(3x+y+z)(9x2+y2+z2-3xy-3yz-3xz)

Answer 2

Answer:

27x 3 + y 3 + z 3 − 9xyz

= (3x)3 + (y)3 + z 3 − 3(3x)(y)(z)

= (3x + y + z) (3x)2 + (y)2 + (z)2 − (3x)(y) − (y)(z) − (z)(3x)

= (3x + y + z)(9x 2 + y 2 + z 2 − 3xy − yz − 3xz)