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Factor completely.
x, to the power 4 , minus, y, to the power 6
x
4
−y
6

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Attempt 1 out of 5

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Answer 1

To factor the expression \( x^4 - y^6 \) completely, we can recognize it as a difference of squares. The expression can be rewritten as:

\[
x^4 - y^6 = (x^2)^2 - (y^3)^2
\]

Now we can apply the difference of squares formula, which states that \( a^2 - b^2 = (a - b)(a + b) \). Here, \( a = x^2 \) and \( b = y^3 \):

\[
x^4 - y^6 = (x^2 - y^3)(x^2 + y^3)
\]

Next, we can attempt to factor \( x^2 - y^3 \) further, as it is a difference of cubes. Recall that the difference of cubes can be factored using the formula:

\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\]

In this case, we can represent \( y^3 \) as \( (y)(y)(y) \), where \( a = x \) and \( b = y \):

\[
x^2 - y^3 = (x^{\frac{2}{3}} - y)(x^{\frac{4}{3}} + x^{\frac{2}{3}}y + y^2)
\]

However, for \( x^2 + y^3 \), we cannot factor it further with real numbers.

Putting this all together, the complete factorization of \( x^4 - y^6 \) is:

\[
(x^2 - y^3)(x^2 + y^3) \text{ or } (x - y)(x^2 + xy + y^2)(x^2 + y^3)
\]

Thus, the final answer is:

\[
(x^2 - y^3)(x^2 + y^3)
\]

While we can’t factor \( x^2 + y^3 \) in the same way over the real numbers, the expression \( x^2 - y^3 \) can be factored as \( (x - y)(x^2 + xy + y^2) \).

So, the final factored form is:

\[
(x - y)(x^2 + xy + y^2)(x^2 + y^3)
\]