Question

Factorise

9a^2+1/9a^2-18

Answer 1

$9 {a}^{2} + \frac{1}{9 {a}^{2} } - 18$

$= 9 {a}^{2} + \frac{1}{9 {a}^{2} } - 2 - 16$

$= (3a)^{2} + ( \frac{1}{3a})^{2} - 2 \times 3a \times \frac{1}{3a} - (4)^{2}$

$= (3a - \frac{1}{3a})^{2} - (4)^{2}$

$= (3a - \frac{1}{3a} + 4)(3a - \frac{1}{3a} - 4)$

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Answer 2

Answer:

(3a -1/3a +4) (3a - 1/3a -4)

Step-by-step explanation:

Above

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