what is the cubed root of 8 squared?
475059
4\ncubed root of 8 is 2. squared is 4. \nOr 8 squared is 64, cubed root of that is 4.
942253
the answer is -3x square root of 2
1033784
A cube has 6 square faces so divide the surface area by six = 81 sq. cm\n\nThe area of a suare is s^2 so so you have to get the suqre root of 81 sq. cm = 9 cm\n\nVolume of a cube is s^3 so raise 9 to the third power  9^3 = 729 cu. cm ---->volume of the cube
240703
1^3 = 1^2\n4^3 = 8^2\n9^3 = 27^2\n\nsee the pattern?  cube a perfect square\n\nso 16^3 = 64^2 will work (4096)\n\n25^3 = 125^2  (you found)\n\n36^3 = 216^2  (46656)\n\ntoo high\n\nSo you only missed 4096.
798797
As you point out, 8 has exactly 3 distinct cube roots.  Only one of these is a real number.  Namely, 2.\n\n( If  you are not comfortable with complex exponentials, I've added a more algebraic method is below )\nExamine the complex exponential a*e^(i*x), where i is the imaginary unit ( a square root of -1 ), and a is a positive real number.  If this is a cube root of 8, then:\n\n( a*e^(i*x) )^3 = a^3 * e^( i * 3x ) = 8 = 8 * e^( i*0 )\n\nFrom this we see that a^3 = 8, so a = 2.\nMoreover, e^( i * 3x ) = e^( i*0 )\nAs e^(z) is 2*pi*i periodic, it must be that\n3x = 2*pi*k for some integer k \n\n = > x = 0, 2*pi/3, 4*pi/3  ( and then repeats )\n\nEuler's Identity says that e^( i * x ) = cos(x) + i * sin(x)\nso e^(i*0) = 1 ( as we already knew )\ne^( i*2*pi/3 ) = cos( 2*pi/3) + i * sin( 2*pi/3)\n                    = (-1/2)  + i * ( sqrt(3) / 2 )\ne^( i*4*pi/3)  = cos( 4*pi/3) + i * sin( 4*pi/3)\n                    = (-1/2)  + i * ( - sqrt(3) / 2 )\n\nMultiplying by the value of a we found above,  the 3 cube roots of 8 are:\n\n2, -1 + i * sqrt(3), and -1 - i*sqrt(3).\n\n\nAn alternate method for cube roots is the following:\n\nThe cube roots of 8 are the solutions of the equation x^3 = 8\nRewriting we find a difference of two cubes which we can factor.\n\nx^3 - 8 = 0\n(x-2)(x^2 + 2x + 4) = 0\nso either\nx - 2 = 0     or    x^2 + 2x + 4 = 0\nx = 2          or    x^2 + 2x + 1 = -3\n                         (x + 1)^2 = -3\n                          x + 1 = +/- i * sqrt(3)\n                          x = -1 +/- i * sqrt(3)\n\nThis method does not work with finding all roots, however.  It works best with cube roots.
769101
= 6 + root(8/2)\n= 6 + root(4)\n= 6 + 2\n= 8
309645
Here : \n\n4^2x3-64 reads 4 squared times the cube root of 64.\n\n= 4x4 x (64)^1/3 = 4 x 4 x 4 = 4^3 = 64\n\nAND \n\n16^7/2 x 2^0 = ((4)^2)^7/2 x 1 = (4)^ (2 x 7/2) = 4^7\n\nCOMPRENDE` ?
912201
Yeah, it is the same as cubed root of 25, but here's how you can go about solving the question to get the answer in simplest form. Cubed root of 5 squared is the same as five squared to the one third (5^2)^1/3\n\n\nMultiplying 2 by 1/3 gives you an exponent of 2/3\nSo your final answer in simplest form is 5 to the 2/3\n\n\nP.S. 8.3 is wrong. Try cubed root of 25 on the calculator, and try five to the two third. They both give you 2.9, not 8.3
1143015
8(8X8+8X8)-8-8-8=1000\n\n-unpredicktaboh-
320716
8(a^4)(b^4) - 28(a^3)(b^3) + 4(a^2)(b^2)\nRemember that (a^N)(b^N) = (ab)^N and let c=ab, then\n8c^4 - 28c^3 + 4c^2 = c^2(8c^2 - 28c +4) or\nc^2(2c^2-7c + 1)/4\n\nAt this point, use the quadratic equation to find the 2 roots from within the parentheses and then make the appropriate substitions for c = ab.
637504
(r+8)^2=r^2+28^2\nor r^2+16*r+64 = r^2+784\nor 16*r+64 = 784\nor r = (784-64)/16 = 720/16 = 45
148359
No true factored form.\n\na=8 b=6 c=8\nx= (-b +- squareroot[b^2-4ac])/2a\nx= (-6 +- squareroot[36-256])/16\nx= (-6 +- squareroot[-220])/16\nx= (-6 +- 2i{squareroot[55]})/16\nx= (-3 +- i{squareroot[55]})/8
740988
Without the correction, your answer would have been -12.  With your correction (cubed rather than squared) your answer should be: -8\n\nHere are the parts:\n-3(12) = -36\n5(-4) = -20\n2 cubed (2^3) = 2 x 2 x 2 = 8\n\nSo it simplifies to:\n-36 - (-20 - 8)\n\nThis further simplifies to:\n-36 - (-28)\n\nWhich finally simplifies to:\n-36 + 28\n= -8\n\nWhew!  Lot's of negatives to keep track of!  Google confirms the answer in the link below:
225517
Well... if I understand your question, it isn't a square root any more. If you have what looks to be a square root sign:\n√¯\nwith (for example) a three over the first little squiggly part, then it would be called a cube root. Any number under the radical sign (root sign) would then have it's cube taken. \n\nFor example, the number 8 under a cube root would be 2, because 2*2*2=8, or 2^3=8.\n\nIf you simply mean that there is a number in front of the square root:\n9√¯\nthis just means that you take the square root and then multiply it by 9. So the number 4 under a square root with a nine in front would be:\n2(the square root of 4)*9=18
771185
Answer-1. \ny=6(cuberoot of x + 2)^2 \n\ndy/dx = 6*2(cuberoot of x + 2)*(1/3)*(x^((1*3) -1)\ndy/dx = 4(x^((1*3) +2)*(x^(-2/3))\ndy/dx(x=8) = 4(8^((1*3) +2)*(8^(-2/3))\ndy/dx(x=8) = 4(2 +2)*(8^(-2/3))\ndy/dx(x=8) = 4(4)*(1/4))\ndy/dx(x=8) = 4\n\nAnswer-2 \ny=x^3+4x^-1 ,x=1\n\ndy/dx = 3*x^2 + 4*(-1)x^(-2) -0\ndy/dx(x=1) = 3*(1^2) +4* (-1)1^(-2)\ndy/dx(x=1) = 3*(1) - 4\ndy/dx(x=1) = -1\n\nI hope thats ok
930340
PROBLEM 1:\n\nsqrt(x^3) = 8\n\nSquare both sides:\nx^3 = 64\n\nNow take the cuberoot of both sides:\nx = cuberoot(64)\n\nSolve:\nx = 4\n\nPROBLEM 2:\n\ncuberoot(x^2) =4 \n\nCube both sides:\nx^2 = 4^3\n\nSimplify:\nx^2 = 64\n\nTake the square root of both sides:\nx = ±8
337853
The answer, before simplification, is sqrt(-8). There is no need to panic, as some of the other answerers did. This simplifies easily to\n\n2 * i * sqrt(2)