How to simplify by factoring (x^3-1)/(x-1)?
I tried (x^2)(x-1)/(x-1), then (x+1)(x-1)(x+1)/(x-1).
We can factor both the sum and the difference of two cubes
Memorize this pattern
(A^3 + b^3) = A^2 - AB + b^2)
(A^3 - b^3) = A^2 + AB + b^2)
(x^3-1)/(x-1)
= (x-1)(x^2 + x + 1)/(x-1)
= x^2 + x + 1, x ≠ 1
Apply difference of cubes formula:
x³ − y³ = ( x − y ) ( x² + x ∙ y + y² )
Replace y with 1
x³ − 1³ = ( x − 1 ) ( x² + x ∙ 1 + 1² )
x³ − 1 = ( x − 1 ) ( x² + x + 1 )
So:
( x³ − 1) / ( x - 1 ) = ( x − 1 ) ∙ ( x² + x + 1 ) / ( x - 1 ) = x² + x + 1
You also can use long division of polynomials.
Result is again:
( x³ − 1) / ( x - 1 ) = x² + x + 1
To simplify the expression (x^3-1)/(x-1) by factoring, we can use the difference of cubes formula.
The difference of cubes formula states that for any two numbers a and b, we have:
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
In our expression, we have x^3 - 1, which can be written as (x)^3 - 1^3. Comparing this to the difference of cubes formula, we can see that a = x and b = 1.
So, using the difference of cubes formula, we can rewrite the expression as:
(x^3 - 1) = ((x)^3 - 1^3) = (x - 1)(x^2 + x + 1)
Now we can simplify the expression (x^3-1)/(x-1):
(x^3 - 1)/(x - 1) = [(x - 1)(x^2 + x + 1)]/(x - 1)
Next, we can cancel out the common factor of (x - 1) in the numerator and denominator:
= (x^2 + x + 1)
Therefore, the simplified expression is x^2 + x + 1.
Your attempt to simplify the expression by factoring was a good start, but there was a small mistake in the second step.
You correctly factored x^3 - 1 as (x - 1)(x^2 + x + 1), but then you canceled out the common factor (x - 1) in the next step. However, you also canceled out the incorrect factor (x - 1) in the denominator, resulting in an incorrect final expression.
Remember, when canceling out common factors, it is important to ensure that the factor being canceled is present in both the numerator and the denominator.