factor y^3+27?
回答 (6)
2008-07-16 3:48 pm
✔ 最佳答案
This is the sum of two cubes:(y³ + 3³) = (y + 3)(y² - 3y + 9)
2008-07-16 10:50 pm
Since both terms are perfect cubes, the binomial can be factored using the sum of cubes formula:
(y+3)(y2-3y+9)
Try this math site it will show you step by step how to do the problem!!
http://mathway.com/problem.aspx?p=basicmath
(y+3)(y2-3y+9)
Try this math site it will show you step by step how to do the problem!!
http://mathway.com/problem.aspx?p=basicmath
2008-07-16 10:49 pm
x³+y³ = (x+y)(x²-xy+y²)
y³+3³ = (y+3)(y²-3y+9)
y²-3y+9 has no real roots.
(3±â9-36)/2 = 3/2 ± (â-27)/2 = (3±3iâ3)/2
Factors: y=-3, (3+3iâ3)/2, (3-3iâ3)/2
y³+3³ = (y+3)(y²-3y+9)
y²-3y+9 has no real roots.
(3±â9-36)/2 = 3/2 ± (â-27)/2 = (3±3iâ3)/2
Factors: y=-3, (3+3iâ3)/2, (3-3iâ3)/2
2008-07-16 10:49 pm
use formula
a^3+b^3=(a+b)(a^2-ab+b^2)
y^3+27
=y^3+ 3^3
=(y+3)(y^2-3y+9) ans
a^3+b^3=(a+b)(a^2-ab+b^2)
y^3+27
=y^3+ 3^3
=(y+3)(y^2-3y+9) ans
2008-07-16 10:48 pm
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
y^3 + 27
= (y + 3)(y^2 - 3y + 9)
y^3 + 27
= (y + 3)(y^2 - 3y + 9)
2008-07-16 10:48 pm
1) cube root each term, place in first brackets.
2)
a) square the first
b) negative product
c) square the second.
y^3 + 27
(y + 3) (? + ? + ?)
"Square the first" (square y)
(y + 3)(y^2 + ? + ?)
"Negative product" (of 3 and y).
Take the product of 3 and y, and then negate the result by multiplying it by (-1).
(y + 3)(y^2 - 3y + ?)
"Square the second" (square 3)
(y + 3)(y^2 - 3y + 9)
2)
a) square the first
b) negative product
c) square the second.
y^3 + 27
(y + 3) (? + ? + ?)
"Square the first" (square y)
(y + 3)(y^2 + ? + ?)
"Negative product" (of 3 and y).
Take the product of 3 and y, and then negate the result by multiplying it by (-1).
(y + 3)(y^2 - 3y + ?)
"Square the second" (square 3)
(y + 3)(y^2 - 3y + 9)
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