Question

If $A={\left(4+\sqrt{15}\right)}^{1/3}+{\left(4-\sqrt{15}\right)}^{1/3}$ , then $A^3-3A$ is equal to

Answer 1

$A=(4+\sqrt{15}{)}^{1/3}+(4-\sqrt{15}{)}^{1/3}$

$A^3=4+\sqrt{15}+4-\sqrt{15}+3(4+\sqrt{15}{)}^{1/3}(4-\sqrt{15}{)}^{1/3}\left(\right(4+\sqrt{15}{)}^{1/3}+(4-\sqrt{15}{)}^{1/3})$

$A^3=8+3(16-15{)}^{2/3}A⟹A^3=8+3A$

$A^3-3A=8$