Question

Factorise 27x³+y³+2³+9xyz

Answer 1

sorry you put + in 9xyz plzz there put —.

It's OK

The Solution is :

USING IDENTITY

x³ + y³ + z³ —3xyz (x+y+z) (x²+y²+z²—xy—yz—zx)

(3x)³ +(y)³ +(z)³ —3 X 3x X y X z (3x+y+z) [(3x)² +(y)² +(z)² —3x X y —y X z —z X 3x]

(3x+y+z) [9x² +y² +z² —3xy —yz —3zx]

I hope u get this

THANKS