Tuesday

November 25, 2014

November 25, 2014

Posted by **parul** on Tuesday, May 29, 2012 at 10:02pm.

a)8a^3+b^3+12a^2 b+6ab^2.

b)27-125a^3-135a+225a^2.

c)27a^3-1/216-9a^2/2+a/4.

d)x^3-1/x63-3x+3/x.

plz help me till 2 o'clock...

- math -
**MathMate**, Tuesday, May 29, 2012 at 10:31pm8a^3+b^3+12a^2 b+6ab^2

Factor into a cube using the identity:

(x+y)^3 = x^3+3x^2y+3xy^2+y^3

a)8a^3+b^3+12a^2 b+6ab^2.

can be factorized into (2a+b)^3

for b)27-125a^3-135a+225a^2.

note the terms 27 and -125a^3 are perfect cubes, which gives you the hints

27=3^3, -125a^3=(-5a)^3

27a^3-1/216-9a^2/2+a/4.

again, 27a^3=(3a)^3, and 1/216=(1/6)^3

d)x^3-1/x63-3x+3/x.

is probably a typo, should read:

d)x^3-1/x^3-3x+3/x.

Again, use -1/x^3=(-1/x)^3 and x^3

now use the identity.

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