Find the units digit in the expansion of $76^{3} + 54^{3}$

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Solution

The correct option is B
0

The expansion of $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$
Keeping a =76 and b =54

We get a + b = 76 + 54

= 130

Multiple of any number with zero in its units place has its unit digit as 0.

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