Question

Without finding cubes, factorise and find the value of (1/4)whole cube + (1/3)whole cube-(7/12)whole cube

Answer 1

when a+b+c=0 then a^3+b^3+c^3=3abc
so, answer is -7/48.
Formula: (a)^3+(b)^3+(c)^3-(3abc)=(a+b+c)((a)^2+(b)^2+(c)^2-ab-bc-ca)

Answer 2

$as \: you \: can \: see \: that \: in \: this \: question \: the \: sum \: of \: all \: numbers \: that \: are \: \binom{1}{4} + \binom{1}{3} - \binom{7}{12} are \: 0$
for the answer if this question it has a identity where a+b+c =0 then a^3 + b^3 + c^3 = 3abc

answer is
$3 \times \binom{1}{4} \times \binom{1}{3} \times \binom{ - 7}{12}$
$= \binom{-7}{48}$
hope this helps you
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