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Pls solve question 7

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Let n, n + 1, n + 2 be three consecutive natural numbers . We have to show that n^{3} + {(n + 1)}^{3} + {(n + 2)}^{3} is divisible by 9 . For n = 1 : 1^{3} + 2^{3} + 3^{3} = 1 + 8 + 27 = 36 which is divisible by 9 . Assume that P (k) is true . ie k^{3} + {(k + 1)}^{3} + {(k + 2)}^{3} is divisible by 9 . Then, we have, = [{(k + 1)}^{3} + {(k + 2)}^{3}] + [k^{3} +^{3} C_{1} k^{2} (3) +^{3} C_{2} k {(3)}^{2} +^{3} C_{3} {(3)}^{3}] = {(k + 1)}^{3} + {(k + 2)}^{2} + k^{3} + [9 k^{2} + 27 k + 27] = [k^{3} + {(k + 1)}^{3} + {(k + 2)}^{3}] + 9 [k^{2} + 3 k + 3] k^{3} + {(k + 1)}^{3} + {(k + 2)}^{3} is divisible by 9 and 9 [k^{2} + 3 k + 3] is divisible by 9 . \Rightarrow {(k + 1)}^{3} + {(k + 2)}^{3} + {(k + 3)}^{3} is divisible by 9 . Hence assuming P (k) is true, P (k + 1) is also true . Therefore, P (n) is true for a l l n \geq 1 .

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