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Test Your Knowledge

• �Factorise : 27x3 + y3 + z3 – 9xyz
• �If (x + y + z) = 0, Show that x3 + y3 + z3 = 3xyz

Verify with me

• �27x3 + y3 + z3 – 9xyz
• �: (3x)3 + (y)3 + (z)3 – 3 x (3x) x (y)x (z)
• �: [3x + y + z][(3x)2 + y2 + z2 + - (3x) x (y) – (y) x (z) – (3x) x z
• �: [3x + y + z ] [9x2 + y2 + z2 – 3xy – y z – 3xz],
• � using the identity

Am I Correct?

• �If (x + y + z) = 0, Show that x3 + y3 + z3 = 3xyz
• �
• �(x + y + z) = 0
• �(x + y) = -z
• �(x + y)3 = (-z)3
• �X3 + y3 + 3x2y + 3xy2 = -z3
• �X3 + y3+ 3xy(x + y) = - z3
• �X3 + y3 + 3xy (-z) = -z3
• �X3 + y3 -3xyz = -z3
• �X3 + y3 + z3 = 3xyz

Answer me!!!

• �without actually doing cube find the value,
• � (28)3 + (-15)3 + (-13)3
• �
• �Without actually calculating the cube find (-12)3 + 73 + 53
• �
• �Factorise: 25a2 – 35a + 12
• �
• �35y2 + 13y - 12

Slide 5

Guess & Check

• �25a2 – 35a + 12 = 25a2 – 20a – 15a + 12
• � = 5a(5a – 3) -4(5a – 3)
• � = (5a – 3) (5a – 4)
• �
• �35y2 + 13y – 12 = 35y2 + 28y – 15y – 12
• � = 5y(7y -3) + 4(7y – 3)
• � = (5y + 4)(7y – 3)