Question

If \[a + b + c = 8\] and \[ab + bc + ca = 20\], find the value of \[{a^3} + {b^3} + {c^3} - 3abc\].

Answer 1

Hint: In the given question, we are asked to find the value of \[{a^3} + {b^3} + {c^3} - 3abc\] on the basis of two equations given. We will use both the equations and applicable identities to reach the desired result.
\[{\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca\]
\[{a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - \left( {ab + bc + ca} \right)} \right)\]

Complete step by step solution:
To find:\[{a^3} + {b^3} + {c^3} - 3abc\]
Given, \[a + b + c = 8\]
\[ab + bc + ca = 20\]
Now, we know that,
\[
  {\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca \\
   \Rightarrow {8^2} = {a^2} + {b^2} + {c^2} + 2\left( {20} \right) \\
   \Rightarrow 64 = {a^2} + {b^2} + {c^2} + 40 \\
 \]
Simplifying it, we get,
\[
  {a^2} + {b^2} + {c^2} = 64 - 40 \\
   \Rightarrow {a^2} + {b^2} + {c^2} = 24 \\
 \]
Now, again using the identity\[{a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - \left( {ab + bc + ca} \right)} \right)\],
\[{a^3} + {b^3} + {c^3} - 3abc = \left( 8 \right)\left( {24 - 20} \right)\]
Simplifying it further, we get,
\[ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 32\]

Note: While solving such types of questions, it is important that we should remember all the basic identities and to understand how and where to use these identities.