Third power of 4 is equivalent to :

a⁵ × a⁷ = a^{5 + 7} = a¹²

Third power of 2 = 2³ = 2 × 2 × 2 = 8
And second power of 3 = 3² = 3 × 3 = 9
? Difference = 9 ? 8 = 1

Given equation is
a^-1 - 1
Apply the law of Fractional Exponents and Laws of Exponents
x^-y = 1/x^y
? a^-1 - 1 = 1/a - 1
? a^-1 - 1 = (1 - a)/a
Now divide by a - 1 in above equation,
? ( a^-1 - 1 ) ÷ (a - 1) = (1 - a)/a ÷ (a - 1)
? ( a^-1 - 1 ) ÷ (a - 1) = (1 - a)/a x 1/ (a - 1)
? ( a^-1 - 1 ) ÷ (a - 1) = -1x (a - 1)/a x 1/ (a - 1)
? ( a^-1 - 1 ) ÷ (a - 1) = -1/a
? Required quotient = -1/a.

( 9ⁿ x 3² x (3^-n/2)^-2 - (27)ⁿ ) / ( 3^3m x 2³ ) = 1/27
? ( 3²ⁿ x 3² x 3ⁿ - 3³ⁿ) / ( 3^3m x 2³ ) = 1/27
? ( 3^{2n + n} x 3² - 3³ⁿ) / ( 3^3m x 2³ )= 1/27
? ( 3³ⁿ x 3² - 3³ⁿ) / ( 3^3m x 2³ ) = 1/27
? 3³ⁿ( 3² - 1) / ( 3^3m x 8 ) = 1/27
? 3³ⁿ( 9 - 1) / (3^3m x 8 ) = 1/27
? ( 3³ⁿ x 8 ) / (3^3m x 8 ) = 1/27
? 3³ⁿ / 3^3m = 1/27
? 3^{3n - 3m} = 1/3³
? 3^{3( n - m )} = 3^-3
3(n - m) = -3
or n - m = -1
or m - n = 1

You can square the both the equation and add them to find the answer.

x² + y² = ( (?3 + 1 ) / ( ?3 - 1 ) )² + ( (?3 - 1 ) / (?3 + 1) )²
x² + y² = ( ?3 + 1 )² / ( ?3 - 1 ) ² + (?3 - 1 )² / (?3 + 1) ²
x² + y² = ( ( 3 + 1 + 2 ?3 ) / ( 3 + 1 - 2 ?3 ) ) + ( ( 3 + 1 - 2?3) / ( 3 + 1 + 2 ?3 ) )
x² + y² = ( 4 + 2 ?3 ) / ( 4 - 2 ?3 ) + ( 4 - 2?3 ) / ( 4 + 2 ?3 )
x² + y² = ( 2 + ?3 ) / ( 2 - ?3 ) + ( 2 - ?3 ) / ( 2 + ?3 )
x² + y² = (4 + 3 + 4 ?3 + 4 + 3 - 4 ?3) / ( 4 - 3 )
x² + y² = 14

17^3.5 x 17^7.3 ÷ 17^4.2 = 17^?
Apply the Laws of Exponents ::
(a^m) x (aⁿ) = a^m+n
Fractional Exponents ::
a^m÷ a ⁿ=a^m?n
? 17^{3.5 + 7.3 - 4.2} = 17^?
? 17^6.6 = 17^?
? ? = 6.6

Given that , 289 = 17^x/5
Use the formula if a^x = a ^y then x = y will be true.
? 17^1/2 = 17^x/5
? x /5 = 1/2
? x = 10 .

^L?3 = M^1/L = Surd of L^th order.

Apply the law of Fractional Exponents and Laws of Exponents
(a^m) x (aⁿ) = a^{m + n}
a^-m = 1/a^m
p⁰ = 1
and if p^X = p^Y then X will be equal to Y. means X = Y;
(a^m)ⁿ = a^{m x n}
(a^m) x (aⁿ) = a^{m + n}
a^m ÷ aⁿ = a^{m ? n}
Or
a^m/aⁿ = a^{m ? n}

Given equation is
2^p + 3^q = 17 ----------------(i)
2^{p + 2} - 3^{q + 1} = 5
? 2^p x 2² - 3^q x 3 = 5
? 2^p x 4 - 3q x 3 = 5
? 4 x 2^p - 3 x 3q = 5---------------(ii)
On multiplying Eq. (i) by 3 and adding it, with Eq. (ii), we get

3 x 2^p + 3 x 3^q = 51
4 x 2^p - 3 x 3^q = 5
_____________________________
7 x 2^p = 56

? 7 x 2^p = 7 x 8
? 2^p = 8
? 2^p = 2³
? p = 3

On substituting the value of p in Eq. (i) . we get
2³ + 3^q = 17
? 8 + 3^q = 17
? 3^q = 17 - 8
? 3^q = 9
? 3^q = 3²
? q = 2