(Y^3)+27
2007-01-17 13:16:44 · 6 answers · asked by Deesa 2 in Science & Mathematics ➔ Mathematics
y^3+27
=(y)^3+(3)^3
=(y+3) (y^2-3y+3^2)
=(y+3) (y^2-3y+9)
formula a^3+b^3=(a+b)(a^2-ab+b^2) has been followed
2007-01-17 13:25:18 · answer #1 · answered by alpha 7 · 0⤊ 0⤋
(y^3)+27
= y^3+3^3
= (y+3)(y^2-3y+9)
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If you can't remember the formula of a^3+b^3 = (a+b)(a^2-ab+b^2), you can try long division.
2007-01-17 21:22:09 · answer #2 · answered by sahsjing 7 · 0⤊ 0⤋
The answer is
(y+3)(y^2-3y+9)
2007-01-17 21:27:00 · answer #3 · answered by anyone... 1 · 0⤊ 0⤋
y^2-3y+9
(y+3)\y^3 + 0y^2 + 0y + 27
-(y^3 + 3y^2)
- 3 y^2
-(-3y^2-9y)
+ 9 y + 27
-( 9y + 27)
0
(y+3)(y^2-3y+9)
2007-01-17 21:23:24 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
a^3+b^3= (a+b)(a^2-ab+b^2)
y^3+3^3
(y+3)(y^2-3y+9)
2007-01-17 21:23:53 · answer #5 · answered by 7 · 0⤊ 0⤋
(Y^3)+27
=>
y³ = -27
y = ³\/-27
y = -3
\/ = rt
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2007-01-17 21:34:25 · answer #6 · answered by aeiou 7 · 0⤊ 0⤋