Note first that \(a = 512 = 2^{9}, b = 675 = 3^{3}5^{2}\) and \(c = 720 = 2^{4}3^{2}5,\) and thus \(2c^{2} = 3ab.\)

Next, note that

\(a^{3} + b^{3} + c^{3} = a^{3} + b^{3} - c^{3} + 2c^{2}c = a^{3} + b^{3} - c^{3} + 3abc = (a + b - c)(a^{2} + b^{2} + c^{2} - ab + ac + bc).\)

Thus \(a + b - c = 512 + 675 - 720 = 467\) divides the given expression, i.e., the given expression is composite.
–
Brian Charlesworth
·
2 years, 3 months ago

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TopNewestNote first that \(a = 512 = 2^{9}, b = 675 = 3^{3}5^{2}\) and \(c = 720 = 2^{4}3^{2}5,\) and thus \(2c^{2} = 3ab.\)

Next, note that

\(a^{3} + b^{3} + c^{3} = a^{3} + b^{3} - c^{3} + 2c^{2}c = a^{3} + b^{3} - c^{3} + 3abc = (a + b - c)(a^{2} + b^{2} + c^{2} - ab + ac + bc).\)

Thus \(a + b - c = 512 + 675 - 720 = 467\) divides the given expression, i.e., the given expression is composite. – Brian Charlesworth · 2 years, 3 months ago

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– Calvin Lin Staff · 2 years, 2 months ago

Nice observation with \( 2c^3 = 3ab \)!Log in to reply

512^3-512+675^3-675+720^3-720+(512+675+720) by fermats little theoram it is divisible by 3 – Shiwang Gupta · 2 years, 2 months ago

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– Calvin Lin Staff · 2 years, 2 months ago

Hm, can you explain in detail? I'm pretty sure that the number is not a multiple of 3.Log in to reply

The problem here is the fact that hcf is one. So, I feel that fermats little theorem must be used – Achuthan Raja Venkatesh · 2 years, 3 months ago

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