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.