Question

factorise 27x³+y³+z³-9xyz

Answer 1

27x
3
+y
3
+z
3
−9xyz = (3x)
3
+(y)
3
+(z)
3
−3(3a)(y)(z)
We know identities x
3
+y
3
+z
3
−3xyz=(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx)
Comparing both sides, we get
∴(3x)
3
+(y)
3
+(z)
3
−3(3x)(y)(z)
=(3x+y+z)[(3x)
2
+y
2
+z
2
−3(3x)(y)−3(y)(z)−3(z)(3x)]
=(3x+y+z)(9x
2
+y
2
+z
2
−9xy−yz−

Answer 2

27x3 + y3 + z3 – 9xyz  = (3x)3 + (y)33 + (z)3 – 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 – 3zx)

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