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.

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

a⁵ × a⁷ = a^{5 + 7} = a¹²

Third power of 4 = 4³ = 4 × 4 × 4 = 64

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

Given that, (42 x 229) ÷ (9261)^1/3 = ?
As we know that (9621)^1/3 = (21 x 21 x 21)^1/3 = 21^3x1/3
? 42 x 229 ÷ (21^{3 x} )^1/3 = ?
? 42 x 229 ÷ 21 = ?
? 42 x 229 x 1/ 21 = ?
? 2 x 229 = ?
? ? = 2 x 229
? ? = 458