Fair and Square

Algebra Level 3

Given that a + b = 5 3 \sqrt{a}+\sqrt{b} = 5\sqrt{3} and a b = 49 ab = 49 , find the value of a 3 + b 3 a^3+b^3 .


The answer is 218014.

Clive Chen by Clive Chen , 25, USA

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2 solutions

First note that a + b = ( a + b ) 2 − 2 a b = ( 5 3 ) 2 − 2 49 = 75 − 14 = 61 a + b = (\sqrt{a} + \sqrt{b})^{2} - 2\sqrt{ab} = (5\sqrt{3})^{2} - 2\sqrt{49} = 75 - 14 = 61 .

Then a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) = ( a + b ) ( ( a + b ) 2 − 3 a b ) = 61 × ( 6 1 2 − 3 × 49 ) = 218014 a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2}) = (a + b)((a + b)^{2} - 3ab) = 61 \times (61^{2} - 3 \times 49) = \boxed{218014} .

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Edwin Gray
Mar 2, 2019

[Sqrt(a) + Sqrt(b)]^2 = a + b +2 Sqrt(ab) = a + b +14 = [5*Sqrt(3)]^2 = 75. So a + b = 61. (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = 61^3 = 226981. Then a^3 + b^3 + 3(ab)(a + b) = 226981, and a^3 + b^3 = 226981 - (3)(49)(61) = 218014.

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