How do you factor $36 x^{2} y^{6} - 16$ $36x^2y^6 - 16$ ?

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Answer 1 · 2018-03-10T14:13:54

$4 \cdot (3 x y^{3} + 2) \cdot (3 x y^{3} - 2)$ $4 * (3xy^3 + 2) *(3xy^3 - 2)$

$36 x^{2} y^{6} - 16$ $36x^2y^6-16$

= $((2^{2} + 3^{2} \cdot x^{2}) \cdot y^{6}) - 16)$ $((2^2+3^2*x^2) * y^6) - 16)$

Because $(a^{2} - b^{2})$ $(a^2 - b^2)$ can be factored into $(a + b) \cdot (a - b)$ $(a+b) * (a-b)$

From the simplified expression above, we can factorise the numbers as it is.

Simply half the index numbers as you would normally when you square root a number with an index.

Therefore:
$([(3 x y^{3} + 2) \cdot (3 x y^{3} - 2)] - 16)$ $([(3xy^3 + 2) *(3xy^3 - 2)] - 16)$

We can also square root the 16

So
$4 \cdot (3 x y^{3} + 2) \cdot (3 x y^{3} - 2)$ $4 * (3xy^3 + 2) *(3xy^3 - 2)$

How do you factor 36x^2y^6 - 1636x2y6−1636x^2y^6 - 16?