Question

Without actually calculating the cubes, find the value of each of the following: (i) \[{\left( { - 12} \right)^3} + {\left( 7 \right)^3} + {\left( 5 \right)^3}\] (ii) \[{\left( {28} \right)^3} + {\left( { - 15} \right)^3} + {\left( { - 13} \right)^3}\]

Answer 1

Hint: A cube number of a number is the number obtained when the number is multiplied by itself three times. Cube of a number x can be denoted as \[y = {x^3}\]
It is a very easy task to determine the cube of a number but in this question it is mentioned that we cannot perform simple calculations and so, we need to use the arithmetic identities to get the result.
To solve this question, use cubic identity formula given as: \[{a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)\]
If the left-hand side of an equation is equal to the right side for any values of the variable, then that equation is an identity which is used in algebraic expansions and factorizations. In standard identities, the values are equal to each other regardless of what values are substituted for the variables. In the identity polynomial variables having degree 3 is called cubic identity.

Complete step-by-step answer:
We know that, \[{a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)\]
Re-writing the above identity as:
\[{a^3} + {b^3} + {c^3} = 3abc + \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) - - (i)\]
(i)
\[{\left( { - 12} \right)^3} + {\left( 7 \right)^3} + {\left( 5 \right)^3}\]
Comparing the given expression with equation (i), we can write
\[
  a = - 12 \\
  b = 7 \\
  c = 5 \\
 \]
\[
  {a^3} + {b^3} + {c^3} = 3abc + \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) \\
   = 3\left( { - 12} \right)\left( 7 \right)\left( 5 \right) + \left( {\left( { - 12} \right) + 7 + 5} \right)\left( {{{\left( { - 12} \right)}^2} + {7^2} + {5^2} - \left( {\left( { - 12} \right) \times 7} \right) - 7 \times 5 - \left( {5 \times \left( { - 12} \right)} \right)} \right) \\
   = 3\left( { - 12} \right)\left( 7 \right)\left( 5 \right) + \left( 0 \right)\left( {{{\left( { - 12} \right)}^2} + {7^2} + {5^2} - \left( {\left( { - 12} \right) \times 7} \right) - 7 \times 5 - \left( {5 \times \left( { - 12} \right)} \right)} \right) \\
   = 3\left( { - 12} \right)\left( 7 \right)\left( 5 \right) \\
   = - 1260 \\
 \]
Hence we can say \[{\left( { - 12} \right)^3} + {\left( 7 \right)^3} + {\left( 5 \right)^3} = - 1260\]
(ii)
\[{\left( {28} \right)^3} + {\left( { - 15} \right)^3} + {\left( { - 13} \right)^3}\]
Comparing the given expression with equation (i), we can write
\[
  a = 28 \\
  b = - 15 \\
  c = - 13 \\
 \]
\[
  {a^3} + {b^3} + {c^3} = 3abc + \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) \\
   = 3\left( {28} \right)\left( { - 15} \right)\left( { - 13} \right) + \left( {28 + \left( { - 15} \right) + \left( { - 13} \right)} \right)\left( {{{28}^2} + {{\left( { - 15} \right)}^2} + {{\left( { - 13} \right)}^2} - \left( {28 \times \left( { - 15} \right)} \right) - \left( {\left( { - 15} \right) \times \left( { - 13} \right)} \right) - \left( {28 \times \left( { - 13} \right)} \right)} \right) \\
   = 3\left( {28} \right)\left( { - 15} \right)\left( { - 13} \right) + \left( 0 \right)\left( {{{28}^2} + {{\left( { - 15} \right)}^2} + {{\left( { - 13} \right)}^2} - \left( {28 \times \left( { - 15} \right)} \right) - \left( {\left( { - 15} \right) \times \left( { - 13} \right)} \right) - \left( {28 \times \left( { - 13} \right)} \right)} \right) \\
   = 3\left( {28} \right)\left( { - 15} \right)\left( { - 13} \right) \\
   = 16380 \\
 \]
Hence we can say \[{\left( {28} \right)^3} + {\left( { - 15} \right)^3} + {\left( { - 13} \right)^3} = 16380\]

Note: Students can check whether a given equation is an identity or not by transforming either one side of an equation in such a way that they become equal to the other side of the equation.