If a^{1/3} + b^{1/3} + c^{1/3} = 0 then

a) a+b+c=0

b) (a+b+c)^{3}=27abc

c) a+b+c=3abc

d) a^{3}+b^{3}+c^{3}=0

Also explain how ??

a^{1}^{/3}+ b ^{1/3} + c ^{1/3 }= 0

( a^{1}^{/3} + b^{1}^{/3}) = -c^{1}^{/3}

( a^{1}^{/3} + b^{1}^{/3})^{3} = ( - c^{1}^{/3} )^{3}

( a^{1}^{/3} )^{3} + ( b ^{1/3})^{3} + 3 a^{1}^{/3} . b.^{1/3} ( a ^{1/3} + b^{1}^{/3} ) = - c

a + b + c + 3 a ^{1/3 }b ^{1/3} ( - c ^{1/3} ) = 0

( a + b + c ) = 3 a ^{1/3 }b ^{1/3} c ^{1/3}

( a + b + c )^{3} = ( 3 a ^{1/3 }b ^{1/3} c ^{1/3}) ^{3}

( a + b + c )^{3} = 27 abc

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