Understanding exponents help?

Mathematics is just a game like.

2^2 is 4 and 3^2 is 9. So 4 -- 9 = --5

(--5)^3 = ---5 * --5 * --5 = -- 125 (proved)

Second problem

8 can be written as 2^3 (since 2*2*2 = 8)

So (2^3)^ (--2/3) = 2 ^ (3* --2/3) = 2^(--2) = 1/ (2^2) =1/4 (proved)

Third one.

Hope product of sqrt 8 and sqrt 2 = sqrt (8*2) = sqrt 16 = 4 (proved)

The present problem is that we dont have maths font to present in the apt manner.

For the first one: 2^2 = 4, 3^2 = 9; now you have (4 - 9)^3 :

4 - 9 = -5; -5^3 = -125

The second one: 8^-2/3 8^2 = 64, 64^-1/3 is the same as (1/64^1/3) which = 1/4

The last one: 8 * 2 = 16, the square root is 4

Just incase you don't understand how exponents work: 2^2 is the same as 2*2, while 2^3 is the same as 2*2*2.

2^2 is 4 and 3^2 is 9

9 -4 =5 and 5^3 is 125 <--- answer

plug it in your calculator OR realize this is this same (8^2)^-1/3 which is the same as 1/(8^2)^1/3

1/64^1/3

64^1/3 is the cube root so it's 4 which gives us 1/4.

multiplying the third question together you get square root of 16 which is 4.

Qu 1

( 4 - 9 ) ³ = ( - 5 ) ³ = - 125

Qu 2

1 / 8^(2/3) = 1 / 2 ² = 1 / 4

Qu 3

√8 √2 = 2 √2 √2 = 2 x 2 = 4

(2^2-3^2)^3 = (4-9)^3 = (-5)^3 = -125

8^(-2/3) = (cube root 8)^(-2) = 2^(-2) = (1/2)^2 = 1/4

8^(1/2) * 2^(1/2) = (8*2)^(1/2) = 16^(1/2) = square root 16 = 4

1) 2^2 - 3^2 = 4 - 9 = -5

(-5)^3 = -5 x -5 x -5 = (-1)^3 x 125 = -125

2) 8^(-2/3)

8 = 2^3 => 8^(-2/3) = (2^3)^(-2/3) = 2^(-2) = 1/2^2 = 1/4

3) If i am getting it right you are asking

sqrt(8)*sqrt(2) where sqrt represents square root.

sqrt(8)*sqrt(2) = (2*sqrt(2))*sqrt(2)) as sqrt(8) = 2*sqrt(2)

=2*{sqrt(2)}^2 =2*2 = 4