If a^{3} = 117 + b^{3} and a = 3 + b, then the value of a + b is:

a3 – b^{3} = 117 and a – b = 3 ⇒ a^{3 }– b^{3} = (a – b) (a^{2} + b^{2} +ab) ⇒ 117 = 3 (a^{2} + b^{2} +ab) ⇒ 39 = a^{2} + b^{2} +ab ⇒ a^{2 }+ b^{2} = 39 – ab we Know that (a-b)² = a^{2} +b² - 2ab = (a-b)² = 39 – ab – 2ab = (a-b)² = 39 – 3ab 9 = 39 – 3ab = ab = 10 Now use (a+b)² = a^{2} + b^{2} + 2ab (a+b)² = 39 – ab + 2ab (a+b)² = 39 + ab = 39 + 10 (a+b)² = 49 = a + b = ±7

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