Question

If $a + b + c = 11$ and $ab + bc + ac = 25$, then find the value of ${a^3} + {b^3} + {c^3} - 3abc$
A. 503
B. 505
C. 506
D. 509

Answer 1

Hint: First of all this is a very simple and a very easy problem. This problem deals with solving the given algebraic formula. To solve this problem this formula is applied which is given below:
$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$
The formula of ${\left( {a + b + c} \right)^2}$ is also used here, which is given below:
$ \Rightarrow {\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca$

Complete step-by-step solution:
Given that the sum of the variables $a,b$ and $c$, which is given below:
$ \Rightarrow a + b + c = 11$
Also given that the sum of the product of two variables taken at a time is given by:
$ \Rightarrow ab + bc + ac = 25$
We are asked to find the value of the difference of the sum of the cubes of the variables $a,b,c$and the product of the number 3 and $a,b,c$ together.
We know the formula, which is given below and is applied here.
$ \Rightarrow {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)$
We know the formula of ${\left( {a + b + c} \right)^2}$ , which is given below:
$ \Rightarrow {\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right)$
$ \Rightarrow {\left( {a + b + c} \right)^2} - 2\left( {ab + bc + ca} \right) = {a^2} + {b^2} + {c^2}$
By rearranging the above formula, which is given below:
$\therefore {a^2} + {b^2} + {c^2} = {\left( {a + b + c} \right)^2} - 2\left( {ab + bc + ca} \right)$
Substituting this in right hand side of the formula ${a^3} + {b^3} + {c^3} - 3abc$, which is given below:
$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{{\left( {a + b + c} \right)}^2} - 2\left( {ab + bc + ca} \right) - \left( {ab + bc + ca} \right)} \right)$
$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{{\left( {a + b + c} \right)}^2} - 3\left( {ab + bc + ca} \right)} \right)$
Now substituting the given values of the expressions, in the above formula, as given below:
$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( {11} \right)\left( {{{\left( {11} \right)}^2} - 3\left( {25} \right)} \right)$
$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( {11} \right)\left( {121 - 75} \right)$
On further simplification of the above formula, which is given below:
$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( {11} \right)\left( {46} \right)$
$\therefore {a^3} + {b^3} + {c^3} - 3abc = 506$
The value of ${a^3} + {b^3} + {c^3} - 3abc$ is 506.

Option C is the correct answer.

Note: Please note that here two important algebraic formulas were used to solve this problem. Algebraic formula includes both numbers and letters. Numbers are fixed, i.e. their value is known. Letters or alphabets are used to represent the unknown. There are few more important algebraic formulas which are to be remembered, which are given below:
$ \Rightarrow {\left( {a - b - c} \right)^2} = {a^2} + {b^2} + {c^2} - 2ab - 2bc - 2ca$
$ \Rightarrow {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$
$ \Rightarrow {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$
$ \Rightarrow {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$
$ \Rightarrow {\left( {a + b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right)$