One step you usually want to take is to move any perfect square out of the radical. Example 1, with numbers: root(12) = root(4x3) = root(4) x root(3) = 2 root(3) Example 2, with variables: root(y cubed) = root(y squared times y) = root(y squared) times root(y) = y root(y)
135
Divide both sides of the equation by Y2.That will leave you with X / Y2 = Y* * * * *That doesn't quite do it, does it? Try this:Take the cube root of both sides.y = ³√x.
Joint variation is a variation in which y varies jointly as x or powers of x () and y or powers of z ( ), if there is some nonzero constant k such that , where x ≠ 0, z ≠ 0, and n > 0. Source~http://www.icoachmath.com/SiteMap/JointVariation.html
y^3 + y^2 - 6 (y+2) (y-3) i think
y varies jointly with x and z if: when x is held fixed, y varies with z and when z is held fixed, y varies with x. Bothe x and z may vary together.
y times the cube root of x2y
One step you usually want to take is to move any perfect square out of the radical. Example 1, with numbers: root(12) = root(4x3) = root(4) x root(3) = 2 root(3) Example 2, with variables: root(y cubed) = root(y squared times y) = root(y squared) times root(y) = y root(y)
You can divide each exponent by three; thus, you get: x4/3y4/3 This is equivalent to x (cubic root of x) y (cubic root of y).
√(12x3y3z-2) * √(15xy) = √(180x4y4x-2) = 13.416x2y2z-1 = 13.416(xy)2/z
64 x cubed minus y cubed is (4x - y)(16x^2 + 4xy + y^2)
135
(y + 27)(y^2 - 27y + 729)
It is x^2 + xy + y^2
4x cubed y cubed z divided by x negative squared y negative 1 z sqaured = 4
9y cubed plus 2y squared
(x + y)(x + y)(x + y)