Dear student,Let the two numbers be x and y , where x>ygiven: x3-y3=387alsox33=8⇒x3=83=512now512-y3=387y3=512-387y3=125y=1253=5×5×53y=5cube root of smaller number is 5Regards

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D e a r s t u d e n t, L e t t h e t w o n u m b e r s b e x a n d y, w h e r e x > y g i v e n : x^{3} - y^{3} = 387 a l s o \sqrt[3]{x^{3}} = 8 \Rightarrow x^{3} = 8^{3} = 512 n o w 512 - y^{3} = 387 y^{3} = 512 - 387 y^{3} = 125 y = \sqrt[3]{125} = \sqrt[3]{5 \times 5 \times 5} y = 5 c u b e r o o t o f s m a l l e r n u m b e r i s 5 R e g a r d s

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