Question

Factorise 27 (x-y)cube + 27 (y-z)cube+ (3z - 3x) cube

Answer 1

Step-by-step explanation:

27(x-y)^3+27(y-z)^3+(3z-3x)^3

27(x-y)^3+27(y-z)^3+3^3(z-x)^3

27{(x-y)^3+(y-z)^3+(z-x)^3..............(1)

since, x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)

so, (x-y)^3= x^3-y^3-3x^2y+3xy^2

(y-z)^3= y^3-z^3-3y^2z+3yz^2

and (z-x)^3= z^3-x^3-3z^2x+3zx^2

therefore, put these values in equation (1),we get

27[3{xy^2+yz^2+zx^2-x^2y-y^2z-z^2x}]

81{xy(y-x)+yz(z-y)+zx(x-z)}