Question

Prove that x3+y3+z3-3xyz=(x+y+z)(x2+y2+z2-xy-yz-zx)

Answer 1

Proof:-

We have

$({x}^{3} + {y}^{3} + {z}^{3} - 3xyz) = ( {x}^{3} + {y}^{3}) + {z}^{3} - 3xyz$

$( {x}^{3} + {y}^{3}) - 3xy(x + y) + {z}^{3} - 3xy$

Let (x+y)=u

${u}^{3} - 3xyu + {z}^{3} - 3xyz$

$( {u}^{3} + {z}^{3}) - 3xy(u + z)$

$(u + z)( {u}^{2} - uz +{z}^{2}) - 3xy(u + z)$

$(u + z)( {u}^{2} +{z}^{2} - uz - 3xy)$

$(x + y + z) {(x + y)}^{2} +{z}^{2} - (x + y)z - 3xy$

$(x + y + z)( {x}^{2} +{y}^{2} + {z}^{2} - xy - yz - zx)$

Therefore

$( {x}^{3} +{y}^{3} +{z}^{3} - 3xyz) = (x + y + z)( {x}^{2}+{y}^{2} + {z}^{2} - xy - yz - zx)$

Answer 2

⬆SEE ATTACHMENT⬆

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