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n is the middle integer of three consecutive positive integers.
The three integers are multiplied to give a product.
n is then added to the product.
Prove that the result is a cube number.
n must be used in your proof.
Your final line must consist of only,
= n^3

User Nau
by
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1 Answer

9 votes

Answer :


\boxed{\textsf{ The final answer is \textbf{n}$^{\textbf{3}}$ .}}

Explanation:

Its given that n is the middle out of the three consecutive integers . So ,

The last integer will be :-


\sf\implies Last \ Integer \ = \ n - 1

The next Integer will be :-


\sf\implies Next \ Integer \ = \ n + 1

Now the Question says that the three integers are multipled to give a product . So that would be.


\sf\implies Product_((three\ consecutive\ integers))= (n-1)n(n+1) = (n^2-1)(n) = \pink{n^3-n}

Now thirdly it's given that n is added to the given integer . That would be ,


\sf\implies Adding\ n = \ n^3 - n + n = \pink{n^3}

Here - n and +n gets cancelled. So we are ultimately left out with n³.

Hence the final number is a cube of some number.

User Andrey Pokrovskiy
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