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Use Euclid division leema to show that cube of any positive integer is of the from 9m,9m+1or9m+8

Posted by Kapil Yadav 6 years, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Rashi Kashyap 6 years, 11 months ago

According to Euclid Division Lemma, a=bq+r,b<r<0 a1=3q+0 a2=3q+1 a3=3q+2 a1=3q. (Cubing both side) (a1)3= (3q)3 (a1)3=3q×3q×3q (a1)3=(9)3q3 Let 3q3=m (a1)3=9m (a2)3=3q+1. (Cubing both side) (a2)3=(3q+1)3 (a2)3=(3q)3+2×(3q)2×1+2×3q×(1)2+(1)3 (a2)3=9(3q3+2×(3q)2+6q)+1 let 3q3+2×(3q)2+6q=m (a2)3=9m+1 a3=3q+2. (Cubing both side) (a3)3=(3q+2)3 (a3)3=(3q)3+2×(3q)2×2+2×3q×(2)2+(2)3 (a3)3=9(3q3+2×(3q)2×2+2×3q×(2)2)+8 Let3q3+2×(3q)2×2+2×3q×4=m (a3)3=9m+8 Hence proved

Yogita Ingle 6 years, 11 months ago

<div> <div> <div>

Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

So, we have the following cases :

Case I : When x = 3q.

then, x³ = (3q)³ = 27q³ = 9 (3q³) = 9m, where m = 3q³.

Case II : When x = 3q + 1

then, x³ = (3q + 1)³

= 27q³ + 27q² + 9q + 1

= 9 q (3q² + 3q + 1) + 1

= 9m + 1, where m = q (3q² + 3q + 1)

Case III. When x = 3q + 2

then, x³ = (3q + 2)³

= 27 q³ + 54q² + 36q + 8

= 9q (3q² + 6q + 4) + 8

= 9 m + 8, where m = q (3q² + 6q + 4)

Hence, x³is either of the form 9 m or 9 m + 1 or, 9 m + 8.

</div> </div> </div>

Sujal Gupta 6 years, 11 months ago

Bshshshghdh

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