Question

Which of the following is not a perfect cube?
(a). 2197
(b). 512
(c). 2916
(d). 343

Answer 1

Hint: Take each option and try to find the cube root by elimination method (splitting theorem into groups) or prime factorization method (express number as a multiple of prime number). Thus compare all numbers and find the number which doesn’t have a perfect cube.

Complete step-by-step answer:
A perfect cube is an integer that is equal to some other integer raised to the third power. We refer to raising a number to the third power as cubing the number.
Let us consider the \[{{1}^{st}}\] option 2197. Let us divide it into 2 groups: the first group is 197 and \[{{2}^{nd}}\] group is (2).
In first group 197, the unit digit ends with 7. The cube root of 3 also ends with 7. Thus the unit place of cube root will be 3.
Let us take the second group 2. Taking,
\[{{1}^{3}}=1\] and \[{{2}^{3}}=8\].
Thus \[{{1}^{3}}<2<{{2}^{3}}\]. 1 is smaller, thus 1 comes in the ten’s place.
\[\therefore \] 2197 has a perfect cube root of 13.
Now let us consider the number 512. We can find the cube root with a prime factorization method, where we can find the prime factor of the given numbers. Hence we get the cubes of prime factors. Now on applying the cube root it gets cancelled by the cubed number present within it.
\[\begin{align}
  & \therefore 512=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \\
 & 512=\left( 2\times 2\times 2 \right)\times \left( 2\times 2\times 2 \right)\times \left( 2\times 2\times 2 \right) \\
 & 512={{2}^{3}}\times {{2}^{3}}\times {{2}^{3}} \\
\end{align}\]
\[\begin{align}
  & 2\left| \!{\underline {\,
  512 \,}} \right. \\
 & 2\left| \!{\underline {\,
  256 \,}} \right. \\
 & 2\left| \!{\underline {\,
  128 \,}} \right. \\
 & 2\left| \!{\underline {\,
  64 \,}} \right. \\
 & 2\left| \!{\underline {\,
  32 \,}} \right. \\
 & 2\left| \!{\underline {\,
  16 \,}} \right. \\
 & 2\left| \!{\underline {\,
  8 \,}} \right. \\
 & 2\left| \!{\underline {\,
  4 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2 \,}} \right. \\
 & \left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
Thus using the law of exponent, \[{{x}^{m}}.{{x}^{n}}={{x}^{m+n}}\].
\[\therefore 512={{2}^{3+3+3}}={{2}^{9}}\]
We can also write, \[{{2}^{9}}={{\left( {{2}^{3}} \right)}^{3}}={{8}^{3}}\], i.e. by law of exponent \[{{\left( {{x}^{m}} \right)}^{n}}={{x}^{mn}}\].
Thus we got the perfect cube of 512 as 8.
Now let’s get the cube root of 2916, by using prime factorization.
Thus 2916 becomes,
\[\begin{align}
  & 2916=2\times 2\times 3\times 3\times 3\times 3\times 3\times 3 \\
 & 2916=\left( 3\times 3\times 3 \right)\times \left( 3\times 3\times 3 \right)\times 2\times 2 \\
 & 2916={{3}^{3}}\times {{3}^{3}}\times 2\times 2=9\sqrt[3]{4} \\
\end{align}\]
\[\begin{align}
  & 2\left| \!{\underline {\,
  2916 \,}} \right. \\
 & 2\left| \!{\underline {\,
  1458 \,}} \right. \\
 & 3\left| \!{\underline {\,
  729 \,}} \right. \\
 & 3\left| \!{\underline {\,
  243 \,}} \right. \\
 & 3\left| \!{\underline {\,
  81 \,}} \right. \\
 & 3\left| \!{\underline {\,
  27 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3 \\
\end{align}\]
Thus from this we can say that 2916 doesn’t have a perfect cube.
Now let us find the cube root of 343 by using factorization.
\[\begin{align}
  & 343=7\times 7\times 7 \\
 & 343={{7}^{3}} \\
\end{align}\]
\[\begin{align}
  & 7\left| \!{\underline {\,
  343 \,}} \right. \\
 & 7\left| \!{\underline {\,
  49 \,}} \right. \\
 & 7\left| \!{\underline {\,
  7 \,}} \right. \\
 & 1 \\
\end{align}\]
Thus 343 has a perfect cube of 7.
Now let us make a summary of what we got.
\[\begin{align}
  & 2197=\sqrt[3]{2197}=13 \\
 & 512=\sqrt[3]{512}=8 \\
 & 2916=\sqrt[3]{2916}=9\sqrt[3]{4} \\
 & 343=\sqrt[3]{343}=7 \\
\end{align}\]
Now looking at the above we can say that the number is 2197, 512 and 343 has a perfect cube. But 2916 doesn’t have a perfect cube.
\[\therefore \] 2916 doesn’t have a perfect cube.
\[\therefore \] Option (c) is the correct answer.

Note: To calculate large numbers, the estimation method is easier. It is the grouping of the numbers from the right. We did an estimation of 2197. For smaller numbers like 512 and 343 prime factorization is much simpler than the estimation method. You can also solve 2197 by prime factorization.
\[2197=13\times 13\times 13\]