Question

factorise:
x³+64y³+z³-12xyz

Answer 1

Answer:

x+y+z whol cube is the answer

Answer 2

SOLUTION

TO FACTORISE

$\sf{ {x}^{3} + 64 {y}^{3} + {z}^{3} - 12xyz }$

FORMULA TO BE IMPLEMENTED

$\sf{ {a}^{3} + {b}^{3} + {c}^{3} -3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) }$

EVALUATION

Here the given expression is

$\sf{ {x}^{3} + 64 {y}^{3} + {z}^{3} - 12xyz }$

We factorise it as below

$\sf{ {x}^{3} + 64 {y}^{3} + {z}^{3} - 12xyz }$

$\sf{ = {x}^{3} + {(4y)}^{3} + {z}^{3} - 3.4x.y.z }$

We know that -

$\sf{ {a}^{3} + {b}^{3} + {c}^{3} -3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) }$

We take a = x , b = 4y , c = z

Thus we get

$\sf{ {x}^{3} + 64 {y}^{3} + {z}^{3} - 12xyz }$

$\sf{= (x + 4y+ z)( {x}^{2} +16 {y}^{2} + {z}^{2} - 4xy - 4yz - xz) }$

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