If a and b are natural numbers satisfying a3 – b3 = ab + 61, then find the value of a + b.
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Answer:
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Step-by-step explanation:
If a − b is divisible by 3, then a − b = 3k, for some integer k
(a − b)² = (3k)²
a² − 2ab + b² = 9k²
a³ − b³ = (a−b) (a² + ab + b²)
= (a−b) (a² − 2ab + b² + 3ab)
= 3k (9k + 3ab)
= 3k * 3 (3k + ab)
= 9 k(3k+ab)
Since k(3k+ab) is an integer, then 9k(3k+ab) is divisible by 9
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Answered by
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Answer:
Step-by-step explanation:
a^3 - b^3 = ( a-b) ( a^2 + ab + b^2) = ab + 61
(a-b )^3 = a^3 - b^3 - 3ab(a-b) = ab + 61
putting value for a^3 - b^3
= ab + 61 - 3ab ( a-b)
-2ab + 61 = a-b
a = 5 , b=6
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