Posted by parul on Tuesday, May 29, 2012 at 10:02pm.
8a^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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