Answer :

We know that, We can express a number which is not a multiple of 3 as ( 3m + 1) or ( 3m +2).

( 3m+1)^{2} = (3m)^{2} + 2×3m×1 + 1^{2} = 9m^{2} + 6m + 1 = 3 ( 3m^{2} + 2m) + 1 ………..(1)

…………..using, (x+y)^{2} = x^{2} + 2xy + y^{2}

( 3m+2)^{2} = (3m)^{2} + 2×3m×2 + 2^{2} = 9m^{2} + 12m + 4 = 9m^{2} + 12m + 3 +1

= 3( 3m ^{2} + 4m +1) +1 ………..(2)

…………..using, (x+y)^{2} = x^{2} + 2xy + y^{2}

Now, observe (1) and (2) carefully, clearly on division by 3 both these terms will leave remainder 1.

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<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I