Q: What is the square root of -1 to the power of 3?

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The square root of 1/9 is 1/3 because the square root of 1 is 1 and the square root of 9 is 3.

The fraction in an exponent takes the base to that radical power (1/2 is take the square root, 1/3 is take the third or "cube" root, 1/5 is take the fifth root, etc) and the numerator to that positive power (2 is squared, 3 is cubed, etc) For example: 4 to the 1/2 power is the square root of 4, which is two. 4 to the 3/2 power is the square root of four, 2, taken to the third power, which is 8. Sources: Background Knowledge

sqrt 3/sqrt3 = 1

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It's the same

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( (sq. root ( 3 ) ) * ( 1 / 9 ) ) cubed

You divide the exponent by 2 The square root of a number is that number to the 1/2 power So for examle square root of x cubed (x^3) is x^3 to the half power =x^3/2 (x to the 3/2 power)

The square root of 1/9 is 1/3 because the square root of 1 is 1 and the square root of 9 is 3.

9 times the square root of 3

rearrange the following: A^(1/n)= the nth root of A. eg A to the power 1/2 equals the square root of A. A to the power 1/3 equals the cube root of A. etc.

4^3√32-3^3√108+^3√256

The fraction in an exponent takes the base to that radical power (1/2 is take the square root, 1/3 is take the third or "cube" root, 1/5 is take the fifth root, etc) and the numerator to that positive power (2 is squared, 3 is cubed, etc) For example: 4 to the 1/2 power is the square root of 4, which is two. 4 to the 3/2 power is the square root of four, 2, taken to the third power, which is 8. Sources: Background Knowledge

Calculate the square root of 512 on any calculator, then multiply the result by 3.Note: If you actually mean the third root, this is NOT called the square root. You can calculate this in Excel with the formula: =512^(1/3) In other words, raise 512 to the power 1/3.

The given term is 3√-1. Since √-1 = i, we can rewrite the term as 3i.

x + y = 1xy = 1y = 1 - xx(1 - x) = 1x - x^2 = 1-x^2 + x - 1 = 0 or multiplying all terms by -1;(-x^2)(-1) + (x)(-1) - (1)(-1) = 0x^2 - x + 1 = 0The roots are complex numbers. Use the quadratic formula and find them:a = 1, b = -1, and c = 1x = [-b + square root of (b^2 - (4)(a)(c)]/2a orx = [-b - square root of (b^2 - (4)(a)(c)]/2aSox = [-(-1) + square root of ((-1)^2 - (4)(1)(1)]/2(1)x = [1 + square root of (1 - 4]/2x = [1 + square root of (- 3)]/2 orx = [1 + square root of (-1 )(3)]/2; substitute (-1) = i^2;x = [1 + square root of (i^2 )(3)]/2x = [1 + (square root of 3)i]/2x = 1/2 + [i(square root of 3]/2 andx = 1/2 - [i(square root of 3)]/2Since we have two values for x, we will find also two values for yy = 1 - xy = 1 - [1/2 + (i(square root of 3))/2]y = 1 - 1/2 - [i(square root of 3)]/2y = 1/2 - [i(square root of 3)]/2 andy = 1 - [1/2 - (i(square root of 3))/2)]y = 1 - 1/2 + [i(square root of 3))/2]y = 1/2 + [i(square root of 3)]/2Thus, these numbers are:1. x = 1/2 + [i(square root of 3)]/2 and y = 1/2 - [i(square root of 3)]/22. x = 1/2 - [i(square root of 3)]/2 and y = 1/2 + [i(square root of 3)]/2Let's check this:x + y = 11/2 + [i(square root of 3)]/2 +1/2 - [i(square root of 3)]/2 = 1/2 + 1/2 = 1xy = 1[1/2 + [i(square root of 3)]/2] [1/2 - [i(square root of 3)]/2]= (1/2)(1/2) -(1/2)[i(square root of 3)]/2] + [i(square root of 3)]/2](1/2) - [i(square root of 3)]/2] [i(square root of 3)]/2]= 1/4 - [i(square root of 3)]/4 + [i(square root of 3)]/4 - (3i^2)/4; substitute ( i^2)=-1:= 1/4 - [(3)(-1)]/4= 1/4 + 3/4= 4/4=1In the same way we check and two other values of x and y.

Cube root is represented as to the power of 1/3.Squared is represented as to the power of 2.A cube root squared is therefore equal to the power of 2 x 1/3 = 2/3.

x2+3i=0 so x2=-3i x=square root of (-3i)=square root (-3)square root (i) =i(square root(3)([1/(square root (2)](1+i) and i(square root(3)([-1/(square root (2)](1+i) You can multiply through by i if you want, but I left it since it shows you where the answer came from. Note: The square root of i is 1/square root 2(1+i) and -1/square root of 2 (1+i) to see this, try and square them!