Question

How do you factor \[8{x^3} - 8\] ?

Answer 1

Hint: To solve this question, first pull out the same terms from the equation and reduce the equation. Then by using the difference of cubes formula, factorize the given equation and equate each term in the equation to zero separately. Then we can find the factors of $x$ directly or by using the Quadratic formula.

Complete step by step answer:
From the given equation: $8{x^3} - 8 = 0..........\left( 1 \right)$
Pull out the like terms, so here the like term is $8$
$ \Rightarrow 8({x^3} - 1)..............\left( 2 \right)$
Now, we can rewrite $1$ as ${1^3}$
$ \Rightarrow {x^3} - {1^3}.................\left( 3 \right)$
Since both terms are perfect cubes, we can factor them using the difference of cubes formula
$\left( {{a^3} - {b^3}} \right) = \,\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$
So now we can consider,
$a = x$ and $b = 1$
Substituting the values in the formula, we get
$ \Rightarrow 8\left( {\left( {x - 1} \right)\left( {{x^2} + x \times 1 + {1^2}} \right)} \right).......\left( 4 \right)$
$\therefore $ The Factorization is
$ \Rightarrow \left( {x - 1} \right)\left( {{x^2} + x \times 1 + {1^2}} \right).........\left( 5 \right)$
Now let us try to factorize the $\left( {{x^2} + x + 1} \right)$ term in $\left( 5 \right)$ equation by splitting the middle term
The first term is${x^2}$, its coefficient is $1$ .
The middle term is$ + x$, its coefficient is $1$
The last term "the constant" is $ + 1$
Now multiply the coefficient of the first term by the constant $1 \times 1 = 1$
Next, find two factors of $1$ whose sum equals the coefficient of the middle term, which is $1$.
But we know that the two factors of $1$ is $ - 1$ and$1$. When we add them we will get
$ \Rightarrow - 1 + \left( { - 1} \right) = - 2$
$ \Rightarrow 1 + 1 = 2$
This is not equal to the coefficient of the middle term. Therefore no two such factors can be found and so trinomial functions cannot be factorized.
$\therefore \,8\left( {x - 1} \right)\,\left( {{x^2} + x + 1} \right).........\left( 6 \right)$
Let us equate the $\left( 5 \right)$ equation to zero, to find the solution
$\therefore \,8\left( {x - 1} \right)\,\left( {{x^2} + x + 1} \right) = 0........\left( 7 \right)$
Now separately equate every term to zero
$i)\,8 = 0$[This equation has no solution.]
$ii)\,x - 1 = 0$
$ \Rightarrow x = 1........\left( 8 \right)$
$iii)\,\left( {{x^2} + x + 1} \right) = 0$
According to the Quadratic Formula, the solution for equation of form $A{x^2} + Bx + C = 0$ , where A, B
and C are numbers, often called coefficients, is given by
$x = \,\dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$
Where $A = 1$ ,$B = 1$ and $C = 1$
Substitute the values in the above formula, we get
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt { - 3} }}{2}..........\left( 9 \right)$
In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called
Complex was invented so that negative numbers would have a square root.
Accordingly $\sqrt { - 3} $ can be written as,
$ \Rightarrow \sqrt { - 3} = \,\sqrt {3 \times \left( { - 1} \right)} $
The above term can also be written as
$ \Rightarrow \sqrt 3 \times \sqrt { - 1} $
Where $\sqrt { - 1} $ is equal to $i$
$\therefore \,\sqrt { - 3} = \sqrt 3 \times i$
We know the value of $\sqrt 3 = 1.7321$
$\therefore \left( {\sqrt { - 3} } \right) = 1.7321i$
Now substitute this value in equation $\left( 9 \right)$
$\therefore x = \left( {\dfrac{{ - 1 \pm 1.7321i}}{2}} \right)........\left( {10} \right)$
This equation $\left( 8 \right)$ will have two imaginary solutions
$\, \Rightarrow x = \left( {\dfrac{{ - 1 + 1.7321i}}{2}} \right).......\left( {11} \right)$ and
$ \Rightarrow x = \left( {\dfrac{{ - 1 - 1.7321i}}{2}} \right).........\left( {12} \right)$
Therefore this quadratic equation has totally three solutions $\left( 8 \right)$ ,$\left( {11} \right)$ and $\left( {12} \right)$
$ \Rightarrow x = 1$
$ \Rightarrow \,x = \left( {\dfrac{{ - 1 + 1.7321i}}{2}} \right)$
$ \Rightarrow x = \left( {\dfrac{{ - 1 - 1.7321i}}{2}} \right)$

Note: While solving the above question, students always forget the two main things. They are as follows:
We have to find the factors and not the value of the variable. If you solve using a quadratic formula, you have to find the factors because the formula gives us the values. If you are using the middle term method, you will get the factors directly.
Many students end their question once they have found the values of x or any other variable that you assumed. The question asked the value of y and not the value of the assumed variable. So, in the end, remember to find the values of a given variable.