How do you multiply $(\frac{1}{3} x^{2} - 8) (\frac{1}{3} x^{2} + 8)$ $(\frac { 1} { 3} x ^ { 2} - 8) ( \frac { 1} { 3} x ^ { 2} + 8)$ ?

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Answer 1 · 2017-03-29T21:19:50

$\frac{1}{9} x^{4} - 64$ $1/9x^4-64$

Let's apply the formula:

$(a - b) (a + b) = a^{2} - b^{2}$ $(a-b)(a+b)=a^2-b^2$

Then you would get:

$(\frac{1}{3} x^{2} - 8) (\frac{1}{3} x^{2} + 8) = \frac{1}{9} x^{4} - 64$ $(1/3x^2-8)(1/3x^2+8)=1/9x^4-64$

How do you multiply (13x2−8)(13x2+8)(\frac { 1} { 3} x ^ { 2} - 8) ( \frac { 1} { 3} x ^ { 2} + 8)?