Question

Find the factors of $64{{m}^{3}}-343{{n}^{3}}$

Answer 1

Hint: To solve this question we will first have to take the common values out from the given two terms. And then like that we have to find all the factors until there is no common term left to be taken out. We will use the formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ to factorise it. For that we will first convert 64 in the cubic of 4 and 343 in the cube of 7. And then use the formula to find the factors.

Complete step-by-step answer:

Let’s start solving this question.
It is given that $64{{m}^{3}}-343{{n}^{3}}$,
Now we can write 64 as ${{4}^{3}}$ and 343 as ${{7}^{3}}$
Now we will substitute these value in $64{{m}^{3}}-343{{n}^{3}}$
Now $64{{m}^{3}}-343{{n}^{3}}$ changes to ${{\left( 4m \right)}^{3}}-{{\left( 7n \right)}^{3}}$
Now using the formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ on ${{\left( 4m \right)}^{3}}-{{\left( 7n \right)}^{3}}$ we get,
$\begin{align}
  & =\left( 4m-7n \right)\left( {{\left( 4m \right)}^{2}}+{{\left( 7n \right)}^{2}}+4m\times 7n \right) \\
 & =\left( 4m-7n \right)\left( 16{{m}^{2}}+49{{n}^{2}}+28mn \right) \\
\end{align}$
Hence, the given value has been factorized into $\left( 4m-7n \right)$ and $\left( 16{{m}^{2}}+49{{n}^{2}}+28mn \right)$

Note: To check if the factors are correct or not multiply the two factors and see if it gives us back the real equation. The formula that we have used ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ must be kept in mind and there are many more formulas that helps us to find the factors of the given equation. We just have to convert the equation in the form of formula and then after that it becomes easy to solve.