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3 answers

7y^6 - 28 = 7(y^6 - 4)
but the bracket inside has a difference of squares
= 7(y^3 -2) (y^3+2)

2007-01-22 04:45:26 · answer #1 · answered by doverbeach 3 · 0 0

7y^6-28
=7(y^6 - 4)
=7(y^3 - 2)(y^3 +2)

Both (y^3-2) and (y^3-2) can be further factored but the cube root of 3 would be involved and I assume you only want rational numbers in your factoring.

Forexample, y^3 - 2 = (y + 2^1/3))(y^2 - 2^1/3y + 2^2/3)

2007-01-22 05:04:29 · answer #2 · answered by ironduke8159 7 · 0 1

7y^6 - 28 =

7(y^6 - 4) =

7(y^6 - 2y^3 + 2y^3 - 4

7[y^3(y^6 - 2) + 2(y^3 - 2)

7(y^3 + 2)(Y^3 - 2)

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2007-01-22 05:33:48 · answer #3 · answered by SAMUEL D 7 · 0 1

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