Factorise : 27x³ + y³ + z³ - 9 xyz
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Step-by-step explanation:
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−9xz)
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