Question

How do you factor \[{u^3} - 1\]?

Answer 1

Hint: In this question we have to find the roots from the above quadratic equation by difference of cubes identity. For that we are going to simplify the equation. Next, we extract the square root of both sides and then simplify to arrive at our final answer. And also we are going to add and subtraction in complete step by step solution.

Complete Step by Step Solution:
Since both terms are perfect cubes, factor using the 'difference of cubes' identity:
\[ \Rightarrow {a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})\]
In our example, \[a = u\] and \[b = 1\] :
We get
\[ \Rightarrow {u^3} - 1 = {u^3} - {1^3}\]
Next we use the difference of cubes identity and simplify below equation and we get
\[ \Rightarrow \left( {{u^3} - {1^3}} \right) = (u - 1)({u^2} + u \cdot 1 + {1^2})\]
Now, multiply u by 1 in the right hand side and we get
\[ \Rightarrow \left( {{u^3} - {1^3}} \right) = (u - 1)({u^2} + u + {1^2})\]
Already we know that, one to any power is one.

\[ \Rightarrow \left( {{u^3} - 1} \right) = (u - 1)({u^2} + u + 1)\] This is the required answer.

Note: we have to remember that, for factoring polynomials, "factoring" (or” factoring completely”) is always done using some set of numbers as possible coefficient.
Here, \[{x^2} - 5\] can’t be factored using integer coefficients. (It is irreducible over the integers.)
In fact, in general we find:
\[ \Rightarrow {a^n} - {b^n} = (a - b)({a^n}^{ - 1} + {a^n}^{ - 2}b + {a^n}^{ - 3}{b^2} + ... + {b^n}^{ - 1})\]
Next we multiply the out in the above term. Most of the terms cancel. When n is even, the second factor can be factor further:
\[ \Rightarrow (a - b)({a^n}^{ - 1} + {a^n}^{ - 2}b + {a^n}^{ - 3}{b^2} + ... + {b^n}^{ - 1}) = \left( {a + b} \right)\left( {{a^n}^{ - 2} + {a^n}^{ - 4}{b^2} + {b^n}^{ - 2}} \right)\]
There are no more linear factors with Real coefficients, but there are quadratic factors with Real coefficients.
There is another little hard way to find the answer.
First adding and subtracting ${u^2} + u$and$ - {u^2} - u$.
We get,
\[ \Rightarrow {u^3} - 1 = {u^3} + {u^2} + u - 1{u^2} - u - 1\]
Next, factor by grouping and we get
\[ \Rightarrow {u^3} - 1 = {u^3} + {u^2} + 1u - 1{u^2} - u - 1\]
Now split the above terms and we get
\[ \Rightarrow {u^3} - 1 = u\left( {{u^2} + u + 1} \right) - 1\left( {{u^2} + u + 1} \right)\]
Take the general term in the above equation and we get the final answer
\[ \Rightarrow \left( {{u^3} - 1} \right) = (u - 1)({u^2} + u + 1)\]
This is the required factor.