Question

$(\frac{ {1}^{ - 2} }{3} ) - (\frac{ {1}^{ - 2} }{9} ) \div {3}^{2} = - 8 \\ \\ \\ how$
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Answer 1

$( \frac{1 {}^{ - 2} }{3} ) - ( \frac{1 {}^{ - 2} }{9} ) \div {3}^{2} = - 8$

$\frac{1}{3} - \frac{1}{9} \div 9 = - 8$

$\frac{1}{3} - \frac{1}{9} \times \frac{1}{9} = - 8$

$\frac{1}{3} - \frac{1}{81} = - 8$

$\frac{27 - 1}{81} = - 8$

$\frac{26}{81} \: not \: equal \: to - 8$

Answer 2

Here is the answer

-1/3 - (-1/9) ÷ 9=-8

multiply -1/3 by 3 in both denominator and numerator

-3/9-(-1/9)÷9=-8

-3-1/9÷9=-8

-4/9÷9=-8

9×9/-4=-8

(reciprocal)

81/-4=-8

they aren't equal

hence proven

Hope it helps!