How can i easily solve the question of class 10th maths question the chapter name is real number exercise 1.1. It’s important question of real number. I want to get best solution for this question Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

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# Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

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Let x be any positive integer and y = 3.

By Euclid’s division algorithm, then,

x = 3q+r, where q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Therefore, putting the value of r, we get,

x = 3q

or

x = 3q + 1

or

x = 3q + 2

Now, by taking the cube of all the three above expressions, we get,

Case (i):When r = 0, then,x

^{2}= (3q)^{3}= 27q^{3}= 9(3q^{3})= 9m; where m = 3q^{3}Case (ii):When r = 1, then,x

^{3}= (3q+1)^{3}= (3q)^{3 }+1^{3}+3×3q×1(3q+1) = 27q^{3}+1+27q^{2}+9qTaking 9 as common factor, we get,

x

^{3 }= 9(3q^{3}+3q^{2}+q)+1Putting = m, we get,

Putting (3q

^{3}+3q^{2+}q) = m, we get ,x

^{3}= 9m+1Case (iii): When r = 2, then,x

^{3}= (3q+2)^{3}= (3q)^{3}+2^{3}+3×3q×2(3q+2) = 27q^{3}+54q^{2}+36q+8Taking 9 as common factor, we get,

x

^{3}=9(3q^{3}+6q^{2}+4q)+8Putting (3q

^{3}+6q^{2}+4q) = m, we get ,x

^{3}= 9m+8Therefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.