If a, b, c and d are four odd perfect cube numbers, then which of the following is always a factor of (3โa+3โb)^2 (3โc+3โd)
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a = 1
b = 8
c = 125
d = 343
These are the four odd perfect cube numbers.
∛a=∛1=1,∛b=∛8=2,∛c=∛125=5 and ∛d=∛343=7
(∛a+∛b)^2 (∛c ∛d) will be 192, 360, 512, 576 and 600
(192, 360, 512, 576, 600) = 8
So this means that 8 will always be the factor of (∛a+∛b)^2 (∛c+∛d)
If there is any confusion please leave a comment below.
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