what is the expression

Question

it is cube root of 27h^2s^-7 over cube root of h^3s^6

(^3√27h^2s^-7)/(^3√h^3s^6)

where r, the exponent of h, is:

and t, the exponent of s, is:

Answer 1

You can also use an exponent of 1/3 for the cube root so you can write the problem this way:

(27h²s^-7)^1/3/(h³s⁶)^1/3

Put them together as one cube root, you'll have:

=((27h²s^-7)/(h³s⁶))^1/3

Since 27 is perfect cube, take it outside of the radicand:

= 3((h²s^-7)/(h³s⁶))^1/3

Using the rule dividing powers with the same base, you'll have:

= 3 (h^-1s^-13)^1/3

Using the rule for power of powers, you'll get:

=3 (h^-1/3s^-13/3)

Now to answer your questions:

r = -1/3 and t= -13/3

Answer 2

I assume you want to simplify the expression.

Since you have the cube root in both numerator and denominator, you can extend the cube root so it's over the entire expression (cube root of (27h²s^-7/h³s⁶)

Then, make the s^-7 have a positive exponent by moving it to the denominator. It will combine with the s⁶ that is already there and become s¹³. Now simplify the h²/h³ to become only an "h" in the denominator. That makes the simplified problem the cube root of 27/(h•s¹³)

You can take the cube root of 27, it is "3". You can consider the s¹³ as s¹²•s and take the cube root of the s¹², which is s⁴ (stays in the denominator). That gives you:

Leaving a radical in the denominator is the math equivalent of using "ain't" in an English composition (neither is considered proper). To get rid of the radical in the denominator, multiply both numerator and denominator by the cube root of h²s² which will result in just h•s in the denominator. The "s" can combine with the s⁴ to get s⁵

So the final answer is

3 times the cube root of (h²s²) all divided by (hs⁵)