Question

# Use Euclid division Lemma to show that the cube of any positive integer Is either of the form $$9m,\ 9m+1$$ or, $$9m+8$$ for some integer $$m$$

Solution

## 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 $$x=3q$$In this case, we know$$x^{3}=(3q)^{3}=27q^{3}=9(3q^{3})=9m$$, where $$m=3q^{3}$$Case $$II$$, When $$x=3q+1$$In this case, we have$$x^{3}=(3q+1)^{3}$$$$\Rightarrow \quad x^{3}=27q^{3}+27q^{2}+9q+1$$$$\Rightarrow \quad x^{3}=9q(3q^{2}+3q+1)+1$$$$\Rightarrow \quad x^{3}=9m+1$$, where $$m=q(3q^{2}+3q+1)$$Case $$III$$, When $$x=3q+2$$In this case, we have$$x^{3}=(3q+2)^{3}$$$$\Rightarrow \quad x^{3}=27q^{3}+54q^{2}+36q+8$$$$\Rightarrow \quad x^{3}=9q(3q^{2}+6q+4)+8$$$$x^{3}=9m+8$$, where $$m=q(3q^{2}+6q+4)$$Hence $$x^{3}$$ is either of the form $$9m$$ or, $$9m+1$$ or, $$9m+8$$.Mathematics

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