Question

How do you factor ${{x}^{6}}-7{{x}^{3}}-8$?

Answer 1

Hint: In this question, we are given an expression in terms of x and we need to factorize it. For this, we will first suppose ${{x}^{3}}$ as y and then form a quadratic equation in terms of y. After that we will use the split middle term method to factorize the expression in terms of y. Then we will substitute the value of y as ${{x}^{3}}$ back and get factors. This factors will then further be factored using the formula ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\text{ and }{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$.

Complete step by step solution:
Here we are given the expression in terms of x as ${{x}^{6}}-7{{x}^{3}}-8$.
We need to factorize it. As we can see, the expression is of degree 6 which is not easy to factorize. So for simplification let us substitute ${{x}^{3}}$ as y. The whole expression gets converted in terms of y as ${{x}^{3}}=y\Rightarrow {{x}^{6}}={{y}^{2}}$ (by squaring both sides) so we have the expression as ${{y}^{2}}-7y-8$.
This is a quadratic equation and thus we can factorize it easily. Let us split the middle term method. For the equation $a{{y}^{2}}+by+c$ we want two numbers ${{n}_{1}},{{n}_{2}}$ such that ${{n}_{1}}\cdot {{n}_{2}}=ac\text{ and }{{n}_{1}}+{{n}_{2}}=b$.
Comparing with our equation we want ${{n}_{1}}\cdot {{n}_{2}}=-8\text{ and }{{n}_{1}}+{{n}_{2}}=-7$.
We see that only the numbers -8 and 1 satisfy both conditions as -8+1 = -7 and (-8)(1) = -8.
So splitting the middle term we have ${{y}^{2}}-8y+y-8$.
Taking y common from the first two terms and 1 common from the last two terms we get $y\left( y-8 \right)+1\left( y-8 \right)$.
Now taking (y-8) common from both terms we get $\left( y+1 \right)\left( y-8 \right)$.
Substituting the value of y as ${{x}^{3}}$ back we get $\left( {{x}^{3}}+1 \right)\left( {{x}^{3}}-8 \right)$.
So the given expression has been factored into two factors. Let us factorize it further.
We know that ${{\left( 1 \right)}^{3}}=1$ and 8 can be written as ${{\left( 2 \right)}^{3}}$. So the expression becomes $\left( {{x}^{3}}+{{1}^{3}} \right)\left( {{x}^{3}}-{{2}^{3}} \right)$.
As we can see, the first factor is of the form ${{a}^{3}}+{{b}^{3}}$ which is equal to $\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$. So, ${{x}^{3}}+{{1}^{3}}$ can be written in the same way. Similarly, second factor is of the form ${{a}^{3}}-{{b}^{3}}$ which is equal to $\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$. So ${{x}^{3}}-{{2}^{3}}$ can be written in the same way.
The expression after using identities becomes $\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)$.
This expression cannot be factored further.
Therefore, this is our final factors of the expression ${{x}^{6}}-7{{x}^{3}}-8=\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)$.

Note: While using the identities ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\text{ and }{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ students must take care of the signs. Try to factorize the expression as much as possible. While finding the number ${{n}_{1}},{{n}_{2}}$ take care of signs and make sure numbers fulfill both conditions.