Answer:
Let the three numbers in A.P. be \[a-d,\text{ }a,\text{ }a+d\] Now, \[a-d+a+a+d=12\] \[3a=12\] \[\therefore a=4\] Also, \[{{(4-d)}^{3}}+{{4}^{3}}+{{(4+d)}^{3}}=288\] \[64-48+12{{d}^{2}}-{{d}^{3}}+64+64+48d+12{{d}^{2}}+{{d}^{3}}=288\] \[192+24{{d}^{2}}=288\] \[24{{d}^{2}}=288192\] \[{{d}^{2}}=\frac{96}{24}=4\] \[d=\pm 2\] The numbers are 2, 4, 6 or 6, 4, 2.
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