recall that the product (a + b)(a - b) is the difference of squares, a² - b². examine this same product for…

recall that the product (a + b)(a - b) is the difference of squares, a² - b². examine this same product for radical expressions. choose the product of (3 + √7)(3 - √7) 6 - 3√7 + 3√7 - √14 9 - √21 + √21 - √49 9 - 3√7 + 3√7 - √49 done

recall that the product (a + b)(a - b) is the difference of squares, a² - b². examine this same product for radical expressions. choose the product of (3 + √7)(3 - √7) 6 - 3√7 + 3√7 - √14 9 - √21 + √21 - √49 9 - 3√7 + 3√7 - √49 done

Answer

Explanation:

Step1: Apply difference - of - squares formula

We know that $(a + b)(a - b)=a^{2}-b^{2}$. Here, $a = 3$ and $b=\sqrt{7}$. So, $(3+\sqrt{7})(3 - \sqrt{7})=3^{2}-(\sqrt{7})^{2}$.

Step2: Calculate the squares

$3^{2}=9$ and $(\sqrt{7})^{2}=7$. $3^{2}-(\sqrt{7})^{2}=9 - 7$. Also, if we expand $(3+\sqrt{7})(3 - \sqrt{7})$ using FOIL method: First: $3\times3 = 9$. Outer: $3\times(-\sqrt{7})=-3\sqrt{7}$. Inner: $\sqrt{7}\times3 = 3\sqrt{7}$. Last: $\sqrt{7}\times(-\sqrt{7})=-\sqrt{49}$. Combining these terms: $9-3\sqrt{7}+3\sqrt{7}-\sqrt{49}$.

Answer:

$9 - 3\sqrt{7}+3\sqrt{7}-\sqrt{49}$