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Question
difference of perfect squares (level 2)
this is the only question in this section.
question
factor completely.
$x^6 - y^2$
Step1: Identify the difference of squares
The expression \(x^6 - y^2\) is a difference of two squares. Recall the formula for the difference of squares: \(a^2 - b^2=(a + b)(a - b)\). Here, \(a = x^3\) (since \((x^3)^2=x^6\)) and \(b = y\) (since \(y^2=y^2\)).
Step2: Apply the difference of squares formula
Using the formula \(a^2 - b^2=(a + b)(a - b)\) with \(a = x^3\) and \(b = y\), we get:
\(x^6 - y^2=(x^3 + y)(x^3 - y)\)
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\((x^3 + y)(x^3 - y)\)