Question

How do you factor $ {x^6} - 26{x^3} - 27 $ ?

Answer 1

Hint: In order to completely factorise the above equation, first let the $ t = {x^3} $ so the equation quadratic and factorise the quadratic equation in $ t $ use the Splitting up the middle method .Now put back the value of $ t $ and apply the formula $ {A^3} - {B^3} = \left( {A - B} \right)\left( {{A^2} + {B^2} + AB} \right) $ in the one part and $ {A^3} + {B^3} = \left( {A + B} \right)\left( {{A^2} + {B^2} - AB} \right) $ in the second part to obtained all the factors.

Formula Used:
 $ {A^3} + {B^3} = \left( {A + B} \right)\left( {{A^2} + {B^2} - AB} \right) $
 $ {A^3} - {B^3} = \left( {A - B} \right)\left( {{A^2} + {B^2} + AB} \right) $

Complete step-by-step solution:
We are given an equation $ {x^6} - 26{x^3} - 27 $ .
In order to factorise the above equation,
Let’s assume $ t = {x^3} $ ,out equation now becomes
 $ {t^2} - 26t - 27 $
Now our equation has become quadratic in variable $ t $
Comparing the equation with the standard Quadratic equation $ a{x^2} + bx + c $
a becomes 1
b becomes -26
And c becomes -27
To find factorisation of the quadratic equation we’ll use splitting up the middle term method
So first calculate the product of coefficient of $ {t^2} $ and the constant term which comes to be
 $ = ( - 27) \times 1 = - 27 $
Now the second Step is to find the 2 factors of the number 2 such that the whether addition or subtraction of those numbers is equal to the middle term or coefficient of x and the product of those factors results in the value of constant .
So if we factorize $ - 27 $ , the answer comes to be -27 and 1 as $ - 27 + 1 = - 26 $ that is the middle term . and \[ - 27 \times 1 = - 27\] which is perfectly equal to the constant value.
Now writing the middle term sum of the factors obtained ,so equation becomes
 $
   = {t^2} - 26t - 27 \\
   = {t^2} - 27t + t - 27 \\
   = t\left( {t - 27} \right) + 1\left( {t - 27} \right) \\
   = \left( {t - 27} \right)\left( {t + 27} \right) \\
  $
Putting back the value of $ t $ , we get
 $ = \left( {{x^3} - 27} \right)\left( {{x^3} + 27} \right) $
Since $ 27 = {3^3} $
 $ = \left( {{x^3} - {3^3}} \right)\left( {{x^3} + {3^3}} \right) $
Now applying the formula of difference of cube of two numbers as $ {A^3} - {B^3} = \left( {A - B} \right)\left( {{A^2} + {B^2} + AB} \right) $ to expand the first part and identity of sum of cube of two numbers $ {A^3} + {B^3} = \left( {A + B} \right)\left( {{A^2} + {B^2} - AB} \right) $ in the second part.
 $
   = \left( {x - 3} \right)\left( {{x^2} + {3^3} + \left( x \right)\left( 3 \right)} \right)\left( {x + 1} \right)\left( {{x^2} + {1^3} - \left( x \right)\left( 1 \right)} \right) \\
   = \left( {x - 3} \right)\left( {{x^2} + 9 + 3x} \right)\left( {x + 1} \right)\left( {{x^2} + 1 - x} \right) \\
  $
Hence, we have completely factored our equation.
Therefore, the required answer is $ \left( {x - 3} \right)\left( {{x^2} + 9 + 3x} \right)\left( {x + 1} \right)\left( {{x^2} + 1 - x} \right) $

Note:
1. One must be careful while calculating the answer as calculation error may come.
2.Don’t forget to compare the given quadratic equation with the standard one every time.
3. Write the factors when the middle term is split using the proper sign.
4. Use the formula properly.