y^2/3 - 9y^1/3 + 8 = 0
2006-11-26 13:37:09 · 5 answers · asked by alchem 1 in Science & Mathematics ➔ Mathematics
Let x = y^(1/3)
x^2 - 9x + 8 = 0
(x - 8)(x - 1) = 0
x = 8 or 1
y^(1/3) = 8 or 1
y = 2 or 1
2006-11-26 13:42:39 · answer #1 · answered by Texas Cowgirl 3 · 0⤊ 0⤋
Well the tricky part is that it is a cube root. eliminate that by mulitplying by all numbers with ^3 exponential.
y^(2/3)3 - 9y(1/3)3 + 8^3 = 0
it becomes y^2 - 9y + 512 = 0
therefore solve the quadratic equation. hope that helps.
2006-11-26 21:51:52 · answer #2 · answered by radtadstar 2 · 0⤊ 0⤋
y^(2/3) - 9y^(1/3) + 8 = 0
multiply everything times the 3rd exponent
so you would get y^2 - 9y + 8^3 = 0
and now use quadratric equation
2006-11-26 21:52:36 · answer #3 · answered by Anonymous · 0⤊ 0⤋
You make a substitution for y. let z=y^1/3; then your equation becomes z^2 - 9*z + 8 = 0. This is a simple quadratic you can slove for z; then y will = z^3 to get the final results
2006-11-26 21:42:11 · answer #4 · answered by gp4rts 7 · 1⤊ 0⤋
Let x=y^(1/3), and set the equation in terms of x. It will be quadratic, x^2 + bx + c, for some b and c, which may be positive or negative. Solve for x. Cube that, and you get y.
Finally, check in the original equation.
2006-11-26 21:42:02 · answer #5 · answered by Anonymous · 0⤊ 0⤋