Question

How do you factor $8{{x}^{3}}+{{y}^{3}}$ ?

Answer 1

Hint: For answering this question we have been asked to factorize the given expression $8{{x}^{3}}+{{y}^{3}}$ . We will use the formula for factorization given as ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$. If we observe the given expression is in the form of the left-hand side one. By expressing it as ${{\left( 2x \right)}^{3}}+{{\left( y \right)}^{3}}$ , we can easily factorise it and get the answer.

Complete step-by-step answer:
Now considering from the question we need to factorize the given expression $8{{x}^{3}}+{{y}^{3}}$ .
From the basic concepts, we know that there exists a formula for factorization mathematically given as ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$ .
As we know that the number eight can be written as the cube of two. Hence the given expression can be simply written as
$\Rightarrow 8{{x}^{3}}+{{y}^{3}}={{\left( 2x \right)}^{3}}+{{\left( y \right)}^{3}}$
Here we have a as 2x and b as y. By applying the formulae we will have,
$\Rightarrow {{\left( 2x \right)}^{3}}+{{\left( y \right)}^{3}}=\left( 2x+y \right)\left( 4{{x}^{2}}+{{y}^{2}}-2xy \right)$
We have obtained the factors which cannot be factored further.
Therefore we can conclude that the factors of the given expression $8{{x}^{3}}+{{y}^{3}}$ are $\left( 2x+y \right),\left( 4{{x}^{2}}+{{y}^{2}}-2xy \right)$ .

Note: While answering this question we need to take care of our calculations we perform and concepts that we apply while solving. We must note that we have to convert 8 as ${{2}^{3}}$ to be able to apply the ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$ identity. We cannot apply it for a as $8{{x}^{3}}$ and b as ${{y}^{3}}$ as it will lead to the wrong answer. Similarly, we have another formula for factorization that is mathematically given as ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$ . For example we can factorize the given expression by considering it as $8{{x}^{3}}-{{\left( -y \right)}^{3}}$ and then apply ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$ to factorise further.