Maths help: how do I simplify y^12 - y^3?

y^12 - y^3. Factor out y^3.

y^3(y^9 - 1). Now factor (y^9 - 1) into two binomial equations. This is a difference of two squares.

Answer: y^3(y^3 + 1)(y^3 - 1)

Plots:

Alternate forms:

y^3 (y^9-1)

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

Real roots:

y = 0

y = 1

Complex roots:

y = -(-1)^(1/9)

y = (-1)^(2/9)

y = -(-1)^(1/3)

y = (-1)^(4/9)

y = -(-1)^(5/9)

y = (-1)^(2/3)

y = -(-1)^(7/9)

y = (-1)^(8/9)

Roots in the complex plane:

Polynomial discriminant:

Delta = 0

Derivative:

d/dy(y^12-y^3) = 12 y^11-3 y^2

Indefinite integral:

integral (y^12-y^3) dy = y^13/13-y^4/4+constant

Global minimum:

min {y^12-y^3} = -3/(4 2^(2/3)) at y = 1/2^(2/9)

Definite integral:

integral_0^1(-y^3+y^12) dy = -9/52 ~~ -0.1730769

Definite integral area below the axis between the smallest and largest real roots:

integral_0^1(-y^3+y^12) theta(y^3-y^12) dy = -9/52 ~~ -0.1730769

hope this is what your looking for

its an exp law that when you have the same coffecient to some power and the two values are being subtracted you can subtract the exp.

x^a -x^b = x^(a-b)

y^12 - y^3 = y^(12-3) =y^9

answer = y^9

You can factor out y^3 and write it as:

y^3 (y^9-1)

Then the term in parentheses is the difference of perfect squares, so factoring it gives:

y^3 (y^3 - 1) (y^3+1)

(a million) 3a + 6 -5 = 3a + a million.... (3 x the bracket) then tidy up (2) 4b -12 -3b = b -12 (as above) (3) 6c + 2d + 3c +3-d = 9c +5d (watch this if there's a minus sign between the two phrases)

when you are substracting divide 12/3 =4 the answe is y t the 4th power

y^9

y^3(y^4-1)

y^3(y^2-1)(y^2+1)

y^3(y^2+1)(y+1)(y-1)