Question

question 1-14
factor the expression, $49x^6 - 81y^2$.
\\(\bigcirc\\) \\((7x^3 + 9y)(7x^3 - 9y)\\)
\\(\bigcirc\\) \\((7x^3 - 9y)^2\\)
\\(\bigcirc\\) \\((49x^3 - 81y)(49x^3 + 81y)\\)
\\(\bigcirc\\) \\((49x^3 - 81y)^2\\)

Explanation:

Step1: Identify the form

The expression \(49x^6 - 81y^2\) is a difference of squares. The formula for factoring a difference of squares is \(a^2 - b^2=(a + b)(a - b)\).
Here, \(a^2 = 49x^6\), so \(a=\sqrt{49x^6}=7x^3\) (since \((7x^3)^2 = 49x^6\)). And \(b^2 = 81y^2\), so \(b=\sqrt{81y^2}=9y\) (since \((9y)^2 = 81y^2\)).

Step2: Apply the formula

Using the difference of squares formula \(a^2 - b^2=(a + b)(a - b)\) with \(a = 7x^3\) and \(b = 9y\), we get:
\(49x^6 - 81y^2=(7x^3 + 9y)(7x^3 - 9y)\)

Answer:

\((7x^3 + 9y)(7x^3 - 9y)\)