Write the following cubes in the expanded form : $(3 a+4 b)^{3}$
Write the following cubes in the expanded form : $(3 a+4 b)^{3}$
Comparing the given expression with $(x+y)^{3},$ we find that
$x=3 a$ and $ y=4 b$
So, using Identity $VI$, we have :
$(3 a+4 b)^{3} =(3 a)^{3}+(4 b)^{3}+3(3 a)(4 b)(3 a+4 b) $
$=27 a^{3}+64 b^{3}+108 a^{2} b+144 a b^{2}$
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