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Why a^3+b^3+c^3=3abc

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Why a^3+b^3+c^3=3abc

Posted by Onkar Yadav 8 years, 5 months ago

CBSE > Class 09 > Mathematics

1 answers

Rashmi Bajpayee 8 years, 5 months ago

We know that

{tex}{x^3} + {y^3} + {z^3} - 3xyz = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right){/tex}

Here, if {tex}x+y+z=0{/tex}, then R.H.S. = 0

Then,

{tex}{x^3} + {y^3} + {z^3} - 3xyz = 0{/tex}

=> {tex}{x^3} + {y^3} + {z^3} = 3xyz {/tex}

So, we can say that if {tex}a+b+c=0{/tex}, then {tex}{a^3} + {b^3} + {c^3} = 3abc{/tex}

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