If A=(4+115)13 +(4-V15)', the A’ – 3A is equal to

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( A=(u+1,5)^{1 / 3}+(4 cdot 1 pi)^{1 / 3} ) ( left.A^{3}=(4+115)^{1 / 3}+(4 cdot sqrt{15})^{1 / 3}right]^{3} ) ( A^{3}=4+1 sqrt{5}+4-sqrt{5}+3(4+1 sqrt{5})^{1 / 3}(4-1 / 5) ) ( A^{3}=10+3 A ) ( A^{3}-3 A=16 )

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