We have : \( { (a+b) }^{ 3 } \) = \( { a }^{ 3 } \) + 3 \( { a }^{ 2 } \) b+ 3 a \( { b }^{ 2 } \) + \( { b }^{ 3 } \).
Method : For finding the cube of a two-digit number with the tens digit = a and the units digit b, we make four columns, headed by
\( { a }^{ 3 } \), 3 \( { a }^{ 2 } \) b, 3 a \( { b }^{ 2 } \) and \( { b }^{ 3 } \).
The rest of the procedure is the same as followed in squaring a number by the column method.
We simplify the working as :
Example 1: Find the value of \( { (29) }^{ 3 } \) by the short-cut method.
Solution. Here, a=2 and b=9
Therefore, \( { (29) }^{ 3 } \) = 24389.
Example 2: Find the value of \( { (71) }^{ 3 } \) by the short-cut method.
Solution. Here, a = 7 and b = 1.
Therefore, \( { (71) }^{ 3 } \) = 357911.
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