Question

Find the cubes of the following numbers:
(i) -13
(ii) $3\dfrac{1}{5}$
(iii) $-5\dfrac{1}{7}$

Answer 1

Hint: We first define what we do when we are finding a cube of any number. We show the difference between the integer and fractions. We use the formula of $\dfrac{{{a}^{3}}}{{{b}^{3}}}$ to find the cube. The solution is found by the multiplication of that number with itself twice.

Complete step by step answer:
We need to find the cubes of the given numbers. To find the cube value of any number we need to multiply that number with itself twice. So, for example if we are trying to find the cube value of x, then first we multiply x with x to find $x\times x={{x}^{2}}$, then we multiply again to find ${{x}^{2}}\times x={{x}^{3}}$. So, ${{x}^{3}}$ is the cubic value of x.
For the first problem we need to find the cube of $-13$. The cube will be ${{\left( -13 \right)}^{3}}$.
We multiply it twice with itself.
So, ${{\left( -13 \right)}^{3}}=\left( -13 \right)\left( -13 \right)\left( -13 \right)=169\left( -13 \right)=-2197$.
Cubic value of -13 is -2197.
For the second problem we need to find the cube of $3\dfrac{1}{5}$. We first need to convert the mixed fraction into an improper fraction. Unlike the first case we have a fraction here. The process remains the same. Just the changes are that we need to do the cube of both numerator and denominator and place them in their respective positions.
So, if we are trying to find a cube of $\dfrac{a}{b}$. The cube will be $\dfrac{{{a}^{3}}}{{{b}^{3}}}$.
The fraction is $3\dfrac{1}{5}=\dfrac{16}{5}$
So, the cube will be ${{\left( \dfrac{16}{5} \right)}^{3}}$.
We multiply it twice with itself.
So, ${{\left( \dfrac{16}{5} \right)}^{3}}=\left( \dfrac{16}{5} \right)\left( \dfrac{16}{5} \right)\left( \dfrac{16}{5} \right)=\dfrac{4096}{125}$.
For the third problem we need to find the cube of $-5\dfrac{1}{7}$. We first need to convert the mixed fraction into an improper fraction. Unlike the first case we have a fraction here. The process remains the same. Just the changes are that we need to do the cube of both numerator and denominator and place them in their respective positions.
So, if we are trying to find a cube of $\dfrac{a}{b}$. The cube will be $\dfrac{{{a}^{3}}}{{{b}^{3}}}$.
The fraction is $-5\dfrac{1}{7}=\dfrac{-36}{7}$
So, the cube will be ${{\left( \dfrac{-36}{7} \right)}^{3}}$.
We multiply it twice with itself.
So, ${{\left( \dfrac{-36}{7} \right)}^{3}}=\left( \dfrac{-36}{7} \right)\left( \dfrac{-36}{7} \right)\left( \dfrac{-36}{7} \right)=\dfrac{-46656}{343}$.

Note: The cubic forms of positive and negative digits work following the digit itself. If the value of the digit is negative then the cube value is negative. If the digit is positive then cubic value is positive. This process is opposite to the square where every square value is positive.