Question

How do you factor \[{x^3} - 1000\] ?

Answer 1

Hint: This is a problem related to algebraic identities. This expression given above is of the form \[{a^3} - {b^3}\] such that 1000 is the perfect cube of 10. Thus on expanding this identity we will get the factors also. The factors are nothing but the values of x that satisfy the above expression. So we will first expand it and then find the answer.

Complete step-by-step answer:
Given that
 \[{x^3} - 1000\]
We can simply write 1000 as a cube of 10.
 \[ \Rightarrow {x^3} - {\left( {10} \right)^3}\]
Now we know that \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\] so we can write the above expression as,
 \[ \Rightarrow {x^3} - {\left( {10} \right)^3} = \left( {x - 10} \right)\left( {{x^2} + 10x + {{10}^2}} \right)\]
Thus equating this to zero,
 \[ \Rightarrow \left( {x - 10} \right)\left( {{x^2} + 10x + {{10}^2}} \right) = 0\]
 \[\left( {x - 10} \right) = 0\]
Now taking 10 on right side we get,
 \[ \Rightarrow x = 10\]
Now,
 \[\left( {{x^2} + 10x + {{10}^2}} \right) = 0\]
 \[ \Rightarrow {x^2} + 10x + 100 = 0\]
Using quadratic formula we get
 \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Putting the values,
 \[ = \dfrac{{ - 10 \pm \sqrt {{{10}^2} - 4 \times 1 \times 100} }}{{2 \times 1}}\]
On solving the roots,
 \[ = \dfrac{{ - 10 \pm \sqrt {100 - 400} }}{2}\]
 \[ = \dfrac{{ - 10 \pm \sqrt { - 300} }}{2}\]
taking the root value because 100 is a perfect square of 10.
 \[ = \dfrac{{ - 10 \pm 10\sqrt { - 3} }}{2}\]
Taking 10 common and dividing it by 2 we get,
 \[ = - 5 \pm 5\sqrt 3 i\]
Thus the roots are
 \[ - 5 + 5\sqrt 3 i\& - 5 - 5\sqrt 3 i\]
Thus the factors are \[ \Rightarrow x = 10\] and \[ \Rightarrow x = - 5 + 5\sqrt 3 i\& x = - 5 - 5\sqrt 3 i\] .
So, the correct answer is “ \[ x = 10\] and \[ x = - 5 + 5\sqrt 3 i\& x = - 5 - 5\sqrt 3 i\] .”.

Note: Note that we can’t directly put the factors as \[ \Rightarrow {x^3} = 1000\] and then 10 is the only possible factor. So we took help from algebraic identities. The factors can be imaginary also. When we find the factors of the second factor that is in the form of algebraic expression their roots can be imaginary depending on the value of discriminant. So since the value of discriminant is negative we get imaginary roots.