How to Re-write Square Root in Denominator of Fraction?

Question

I am working an integral problem. I am looking at the solution and a bit confused. Specifically, please take a look at the denominator of the integrand.

I want to re-write the denominator as $(2 + \sec(3P))^{-2/3}$. But the solution has it as $(2 + \sec(3P))^{-1/3}.$

Am I correct (and the solution is wrong) or am I missing something?

Answer 1

That's not a square root in the denominator. It's a cube root.

$\dfrac1{\sqrt[3]a}=\dfrac1{a^{1/3}}=a^{-1/3}.$

Answer 2

$(2 + \sec(3P))^{-2/3}= \frac 1{(2+\sec(3P))^{\frac 23}}= \frac 1{((2+\sec(3P))^2)^{\frac 13}}=\frac 1{\sqrt[3]{(2 + \sec(3P))^2}}$

$(2 + \sec(3P))^{-1/3}= \frac 1{(2+\sec(3P))^{\frac 13}}= \frac 1{\sqrt[3]{2 + \sec(3P)}}$

........

$\sqrt[n] a= a^{\frac 1n}$ so

$\frac 1{\sqrt[3]{2 + \sec(3P)}} = \frac 1{(2+\sec(3P))^{\frac 13}}$

And

$\frac 1a = a^{-1}$ and $\frac 1{a^k} = a^{-k}$ so

$\frac 1{\sqrt[3]{2 + \sec(3P)}} = \frac 1{(2+\sec(3P))^{\frac 13}}= (2+\sec(3P))^{-\frac 13}$.