Expert: Sherman D. - 4/9/2007

Question
i need help simplifying a couple of exquations....

2x^-1(x^2y^3-3y^6)

(4^1/2+25^1/2)^2

(3^-1-4^-1)^-2

(-2^3*16^-1/4)/(4^2)

and how to solve

4(x-3)^2/3=64

Answer
if by these you mean

(2x^-1)(x^2y^3 - 3y^6)
((2x^-1)(x^2y^3)) - ((2x^-1)(3y^6))
(2 * x^(-1 + 2)y^3) - (6x^-1y^6)
(2xy^3) - (6x^-1y^6)
(2xy^3) - ((6y^6)/x)

To make this as one fraction, it would be

(2x^2y^3 - 6y^6)/x

-----------------------------

(4^(1/2) + 25^(1/2))^2
(2 + 5)^2
7^2
49

--------------------------

((3^-1) - (4^(-1)))^(-2)
((1/3) - (1/4))^(-2)
((4/12) - (3/12))^(-2)
((4 - 3)/12)^(-2)
(1/12)^(-2)
12^(2)
144

--------------------------------

((-2^3) * 16^(-1/4))/(4^2)
(-8 * (1/16)^(1/4))/16
(-8 * (1/2))/16
(-8/2)/16
(-4/16)
(-1/4)

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4(x - 3)^(2/3) = 64
(x - 3)^(2/3) = 16
(x - 3)^2 = 4096
x - 3 = -64 or 64
x = -61 or 67