Question

The sum of the cubes of three consecutive natural numbers is divisible by
A. 7
B. 9
C. 25
D. 26

Answer 1

Hint: We will assume any three natural numbers and we will first try to find out the sum of three consecutive natural numbers and then we will use the divisibility test and choose the correct option accordingly. We can start by considering the three consecutive numbers as a - 1, a and a + 1.

Complete step-by-step answer:
It is given in the question that we have to find a number which divides the sum of the cubes of three consecutive natural numbers. Let us assume the three consecutive natural numbers as a - 1, a and a + 1 respectively. Now, it is given that the cubes of three consecutive natural numbers are divisible by some number, which means that we have to find out the cubes of these three consecutive numbers. We have already assumed the three numbers as $\left( a-1 \right),\left( a \right),\left( a+1 \right)$. So, the sum of the cubes of the assumed consecutive numbers will be,
 ${{\left( a-1 \right)}^{3}}+{{\left( a \right)}^{3}}+{{\left( a+1 \right)}^{3}}$
Now, we know that, ${{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}$ and ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$. So, we can write the above relation as,
$\begin{align}
  & {{a}^{3}}-3{{a}^{2}}\times 1+3a\times {{1}^{2}}-{{1}^{3}}+{{a}^{3}}+{{a}^{3}}+3{{a}^{2}}\times 1+3a\times {{1}^{2}}+{{1}^{3}} \\
 & \Rightarrow {{a}^{3}}-3{{a}^{2}}+3a-1+{{a}^{3}}+{{a}^{3}}+3{{a}^{2}}+3a+1 \\
 & \Rightarrow 3{{a}^{3}}+6a \\
\end{align}$
We can take 3 as a common term. So, we get,
$3a\left( {{a}^{2}}+2 \right)$
So, it means that the sum of the cubes of the three consecutive natural numbers is given by $3a\left( {{a}^{2}}+2 \right)$. Now, if we put the value of a = 1 in this obtained relation, then we will get,
$3a\left( {{a}^{2}}+2 \right)\Rightarrow 3\left( 1 \right)\left( {{1}^{2}}+2 \right)\Rightarrow 3\left( 1+2 \right)\Rightarrow 3\times 3=9$
Now, if we put the value of a as 2, then we get,
$3a\left( {{a}^{2}}+2 \right)\Rightarrow 3\left( 2 \right)\left( {{2}^{2}}+2 \right)\Rightarrow 6\left( 4+2 \right)\Rightarrow 6\times 6=36$
And if we put the value of a = 3, then we get,
$3a\left( {{a}^{2}}+2 \right)\Rightarrow 3\left( 3 \right)\left( {{3}^{2}}+2 \right)\Rightarrow 9\left( 9+2 \right)\Rightarrow 9\times 11=99$
So, we get the numbers as 9, 36, 99, …… and if we observe the given options, only option B contains 9 as the number that divides all the three numbers, 9, 36 and 99, respectively. All the rest of the options, that is option A. with 7, option C. with 25 and option D. with 26, do not divide any of the numbers among 9, 36 and 99.
Therefore, option B, 9 is the correct answer.

Note: Most of the time the students make a mistake while assuming the three consecutive numbers. So, we should take them as a - 1, a, a + 1 or as a, a + 1, a + 2 also but it is advisable to take them as a-1,a,a+1 since many terms get cancelled and calculations become easier. Many students also get confused with the use of plus and minus signs in the expansion of the formulas of ${{\left( a+b \right)}^{3}}$ and ${{\left( a-b \right)}^{3}}$ respectively. Therefore, it is advised that the students should be thorough with expansion of these cubic formulas to avoid mistakes.