1
Question
Factorise: 27x3+y3+z3−9xyz
Open in App
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)
Suggest Corrections
4
Similar questions